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Study
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CIRCLEBASIC GEOMETRY WITH CIRCLES
(i) Equal chords subtends equal angles at the centre andvice-versa. fig - (i)
(ii) Equal chords of a circle are equidistant from the centreand vice-versa. fig - (ii)
(iii) Angle subtended by an arc at the centre is double theangle subtended at any point on the remaining part ofthe circle. fig - (iii)
(iv) Angles in the same segment of a circle are equal. fig - (iv)
(v) The sum of the opposite angles of a cyclic quadrilateralis 1800 and vice-versa. fig - (v)
(vi) If a line touches a circle and from the point of contacta chord is drawn, the angles which this chord makeswith the given line are equal respectively to the anglesformed in the corresponding alternate segments. fig - (vi)
(vii) If two chords of a circle intersect either inside oroutside the circle, the rectangle contained by the partsone chord is equal in area to the rectangle containedby the parts of the other.
AP × PB = CP × PD fig - (vii)
(viii) The greater of the two chords in a circle is nearer tothe centre than lesser. fig - (viii)
(ix) A chord drawn across the circular region divides itinto parts each of which is called a segment of thecircle. fig - (ix)
(x) The tangents at the extremities of a chord of a circleare equal. fig - (x)
The angle between the tangents is bisected by the straight line, which joins their pointof intersection to the centre. This straight line also bisects at right angles the chord,which joins the points where they touch the circle
TEACHING NOTES
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LECTURE – 11. DEFINITION : Circle is defined as a locus of a point 'P' which moves in x-y plane in
such a way such that its distance from the fixed point in the same plane is alwaysconstant.i.e. (h – a)2 + (k – b)2 = r2
(x – a)2 + (y – b)2 = r2 ....(1)fixed point is centre of the circle and fixed distance is radius of circle.
a, b centrer radius
Note: If centre is origin then the equation of circle :x2 + y2 = r2
Expand (1) x2 + y2 – 2ax – 2by + a2 + b2 – r2 = 0.hence the general equation of the circle is taken as
x2 + y2 + 2gx + 2fy + c = 0. ....(2)
This can be written as (x + g)2 + (y + f)2 =2
22 cfg
hence, Centre (–g, – f) i.e. ( –2
1 coefficient of x ; –2
1 coefficient of y )
Radius cfg 22 Remember that every second degree equation in x & y in which coeff. of x2=coeff. of y2
and there is no xy term represents a circle.
Note:(i) Necessary and sufficient condition for general equation of degree 2
i.e. ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 to represent a circle is(a) coefficient of x2 = coefficient of y2 (not necessarily unity) and(b) coefficient of xy = 0
(ii) The general equation of circle x2 + y2 + 2gx + 2fy + c = 0 contains 3 independentarbitrary constants g, f and c that a unique circle passes through 3 non-collinear pointshence 3 points on a circle or 3 tangents to a circle or 2 tangents to a circle and a pointetc. must be given to determine the unique equation of the circle.
Nature of circle :x2 + y2 + 2gx + 2y + c = 0
where r = cfg 22
If g2 + f2 – c > 0 If g2 + f2 – c = 0 If g2 + f2 – c < 0 Real circle with finite radius point circle imaginary circle
Note: Always chase for centre and radius to get the equation of circle.
EXAMPLES :Ex-1 Find the equation of the circle(a) through 3 non-collinear points (or to prove that the
points (3, 4), (– 3, – 4), (0, 5) and (–5, 0) are concyclic.
(b) having lines 2x – 3y = 5 and 3x – 4y = 7 as its diameter / Normal / longest chord andwhose area is 154 sq. units.
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(c) which is concentric with 3x2 + 3y2 – 5x – 6y – 14 = 0 and perimeter of its semicircle is 36.
(d) Centre is on the line y = 2x and passing through (–1, 2) and (3, – 2).(e) passes through (2, 3) and centre on the x-axis radius being 5.
(f) which has its centre as the point (4, 3) and touching the line 5x – 12y – 10 = 0.(g) circumscribing the triangle formed by the lines 2x + y = 4 ; x + 2y = 5 ; x + y = 6.
(h) passing through the extremities of the diameter ofthe circle x2 + y2 = 4 ; x2 + y2 + 2x – 4y – 2 = 0and x2 + y2 – 6x – 8y + 10 = 0
[Sol. r2 = a2 + b2 + 4 = (a + 1)2 + (b – 2)2 + 7 = (a – 3)2 + (b – 4)2 + 15]
2. DIAMETRICAL FORM OF CIRCLE :Equation of circle with (x1, y1) and (x2, y2) as its diameter is
(x – x1) (x – x2) + (y – y1) (y – y2) = 0which can be written as
x2 + y2 – x (x1 + x2) – y(y1 + y2) + x1x2 + y1y2 = 0 ....(1)note that x1 , x2 are the roots of the equation
x2 – x (x1 + x2) + x1x2 = 0and y1 , y2 are the roots of the equation
y2 – y (y1 + y2) + y1y2 = 0
Note that Equation (1) is also the equation of the circle with least radius passesthrough (x1, y1) and (x2, y2).
EXAMPLES :Ex-1 The abscissa of 2 points 'A' and 'B' are the roots of the equation x2 + 2x – a2 = 0 and
the ordinate are the roots of the equation y2 + 4y – b2 = 0. Find the equation of circlewith AB as diameter.
Ex-2 Find the equation of a circle circumscribing the quadrilateral formed by the lines2x + 3y = 2 ; 3x – 2y = 4 ; x + 2y = 3 ; 2x – y = 3
Ex-3 Equation of the circle which touches the lines x = 0, y = 0 and x = c will be the circledescribed on the line joining the points (0, c/2) and (c, c/2) as diameter.
[Ans. 4(x2 + y2) – 4cx – 4cy + c2 = 0]
Ex-4 Line y = mx + c cuts the curve y2 = 4ax at A and B.Find the equation of circle with AB as diameter.
3. INTERCEPT (Made by the circle) :"Intercept made by a circle on the coordinate axes", is the distance between the 2points where the circle cuts the axis of x and axis of y.
x2 + y2 + 2gx + 2fy + c = 0For x-intercept : put y = 0
x2 + 2gx + c = 0 ; x1 + x2 = – 2g ; x1 x2 = c
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(i) If g2 – c > 0 circle cuts the x-axis at 2 distinct points.
(ii) If g2 – c = 0 circle touches the x-axis
(iii) If g2 < c circle lies completely above or below the x-axis.
|||ly For y-intercept : put x = 0y2 + 2fx + c = 0
| y1 – y2 | = 2 cf 2
(i) If f2 – c > 0 circle cuts the y-axis at 2 distinct points.
(ii) If f2 = c circle touches the y-axis
(iii) If f2 < c circle lies completely either on right or on left of y-axis.
EXAMPLES :Ex-1 Find the equation of a circle which touches the +ve axis of y at a distance of 4 units
from origin and cuts off an intercept of 6 units from the positive axis of x.[Sol. From figure
equation of circle(x – 5)2 + (y – 4)2 = 52
or (x + 5)2 + (y – 4)2 = 252
Aliter : Let the equation of circle bex2 + y2 + 2gx + 2fy + c = 0
passes through (0, 4) 16 + 8 f + c = 0 ....(1)
x-intercept = 6 2 cg2 = 6 g2 – c = 9 ....(2)
Also touches the y-axis f2 = c ....(3)Now proceed. ]
Ex-2 Find the equation of a circle which touches thecoordinate axes and whose radius = 2.
[Sol. 4 circles are possible(x ± 2)2 + (y ± 2)2 = 22 Ans. ]
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Ex-3 Find the equation of a circle passing throughorigin cutting off intercept equals to unity onthe lines y2 – x2 = 0.
[Sol. Circles described on Ab, BC, CD & DA as diameter.]
Ex-4 Find the equation of the locus of the centre of a circle which touches the positivey-axis and having intercept on x-axis equals to 2l.
[Sol. from figureh2 = k2 + l2
required locus isx2 – y2 = l2 ]
Ex-5 Two rods whose lengths are 2a and 2b slide along the rectangular axes in such a waythat their extremities are always concyclic. Find the equation of the locus of the centreof the circle. [Ans. x2 – y2 = a2 – b2]
H. W. After Lecture-1 : Exercise-17 (S.L. Loney) Leave Q.No. 10, 11, 12, 33
LECTURE – 24. POSITION OF A POINT W.R.T. A CIRCLE :
Point 'P' lies outside / on / inside the circle according asOP r
i.e. (x1 + g)2 + (y1 + f)2 r2
0cfy2gx2yx 1121
21
or S1 0, where S1 = cfy2gx2yx 1121
21
Note: Greatest and least distance of a point A (x1, y1) from a circle with centre 'C' andradius 'r' is
| AC + r | and | AC – r |
EXAMPLES :Ex-1 If the join of (x1, y1) & (x2, y2) makes an obtuse angle at (x3, y3) then prove that
(x3 – x1) (x3 – x2) + (y3 – y1) (y3 – y2) < 0[Sol. Equation of circle with AB as diameter
(x – x1) (x – x2) + (y – y1) (y – y2) = 0since AB subtends obtuse angle at (x3 , y3) P lies inside the circle (x3 – x1) (x3 – x2) + (y3 – y1) (y3 – y2) < 0 H.P. ]
Ex-2 S1 = x2 + y2 – 4x + 6y – 3 = 0S2 = x2 + y2 + 4x – 6y – 3 = 0point (1, 2) lies(A) inside S1 = 0 and inside S2 = 0(B) outside S1 = 0 and outside S2 = 0(C) inside S1 = 0 and outside S2 = 0(D) outside S1 = 0 and inside S2 = 0
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Ex-3 For what value of a the point (a, a + 1) lies inside the regionbounded by the circle x2 + y2 = 4 and the line x + y = 2 inthe first quadrant.
[Ans.
2
17,
2
1]
5. PARAMETRIC EQUATION OF A CIRCLE :
sin
yy
cos
xx 11 = r (Distinguish it with the parametric equation of line, carefully)
x = x1 + r cos
and y = y1 + r sin
x1 y1 fixed centre
r fixed radius and [0, 2) is a parameter.
coordinates of any point on the circle is x1 + 4 cos , y1 + r sinif (x1, y1) (0, 0) x = r cos & y = r sinwhich are the parametric coordinates for the circle x2 + y2 = r2
Note: If is eliminated we get Cartesian form of a circle i.e. (x – x1)2 + (y – y1)
2 = r2.
C.B.S.E.(1) x2 + y2 – 6x + 4y – 3 = 0. convert into parametric form[Sol. x = 3 + 4 cos
y = – 2 + 4 sin ]
(2) If A (cos 1, sin 1) ; B (cos 2, sin 2) ; C (cos 3, sin 3) are the vertices of a triangleABC then find the orthocenter of the triangle.
[Sol. A, B, C lies on a circle x2 + y2 = 1
O (S cos1 ; sin1) Ans. ]
(3) 'P' is the variable point on the circle with centre at C. CA & CB are perpendicular fromC on x-axis and y-axis respectively. Show that the locus of the centroid of triangle PABis a circle with centre at the centroid of triangle CAB and radius equal to the third of theradius of the given circle.
[Sol. Let the circle be x2 + y2 + 2gx + 2fy + c = 0Let the centroid of the CAB be (h, k)
3h = x1 – g + r cosand 3k = y1 – f + r sin
where x1 = – g ; y1 = – f 3h = – 2g + r cos
3k = – 2f + 4 sin (3h + 2g)2 + (3k + 2f)2 = r2 Now proceed ]
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6. LINE AND A CIRCLE :Let L = 0 be a line and S =0 be a circle. If 'r' is the radius of the circle and 'p' is thelength of perpendicular from the centre on the line, then
(i) If p > r line is neither secant nor tangent ; passes outside the circle
(ii) If p = r line is tangent to the circle.(iii) If p < r line is a secant.(iv) If p = 0 line is a diameter.
Alternatively : Solve the line with the circle and if(i) D > 0 line is a Secant.(ii) D = 0 line is a Tangent.(iii) D < 0 line passes outside the circle.
This is true for a line with any 2nd degree curve.
Objective :(1) For what value of 'm' the line 3x – my + 5 = 0 is tangent to the circle
x2 + y2 – 4x + 6y – 3 = 0.[Hint: use p = r ]
(2) If 4l2 – 5m2 + 6l + 1 = 0 then show that the line lx + my + 1 = 0 touches a fixed circle.Find the centre and radius of the circle.
[Sol. Method-I : 4l2 – 5m2 + 6l + 1 = 5m2
Add 5l2 on both sides9l2 + 6l + 1 = 5 (l2 + m2)
(3l + 1)2 = 5 (l2 + m2)
22 m
13
l
l = 5 ....(1)
Equation (1) shows that the line lx + my + 1 touches a fixed circle whose centre is
(3, 0) and radius = 5 .
Method-II : Given 4l2 – 5m2 + 6l + 1 = 0 ....(1)and given line is lx + my + 1 = 0 ....(2)Let line (2) touches the circle whose centre is (, ) and radius is 'a'.
then 22 m
|1m|
l
l = a
(l + m + 1)2 = a2(l2 + m2) (2 – a2) l2 + (2 – a)m2 + 2lm + 2 l + 2m + 1 = 0 ....(3)Comparing (1) and (3) we get
2 – a2 = 42 – a2 = – 52 = 6 = 3
2 = 0 = 0 a = 52 = 0
Hence centre of the required circle is (3, 0) and radius = 5 .
Note: Length of the chord intercepted on the circle by a given line
l = 22 pr
AB = 2l = 2 22 pr H.W. After Lecture - 2 : Nil
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LECTURE – 37. TANGENT AND NORMAL :
Tangent : Tangent is the limiting case of the secant as the point B A
Normal : Normal is a line perpendicular to the tangent passing through the point oftangency. In case of circle normal always passes through centre.
Equation of the tangent drawn to the circle in various forms :(i) Tangent drawn to the circle x2 + y2 + 2gx + 2fy + c = 0 at its point (x1, y1) is given by
xx1 + yy1 + g(x + x1) + f (y + y1) + c = 0
If circle is x2 + y2 = a2
then tangent is xx1 + yy1 = a2 (Cartesian form)
(ii) Parametric form :
sinaycosax
1
1 } equation of tangent is x cos + y sin = a.
(iii) Slope form : y = mx + c is tangent to the circle x2 + y2 = a2
if c2 = a2(1 + m2) condition of tangency.
y = mx ± 2m1a is the equation of the tangent to x2 + y2 = a2.
Note:(i) For a unique value of m there will be 2 tangent which are parallel to each other.(ii) From an external point 2 tangents can be drawn to the circle which are equal in length
and are equally inclined to the line joining the point and the centre of the circle.
(iii) Equation of tangent drawn to any second degree centre at P (x1, y1) on it can beobtained by replacing.x2 x x1 ; y
2 y y1 ; 2x x + x1 ; 2y y + y1 ; 2xy xy1 + yx1
(iv) Point of Tangency :for P : either solve tangent and normal to get Por compare the equation of tangent at (x1, y1) withthe given tangent to get point of tangency.
EXAMPLES :Ex-1 Find the equations of the tangents to the circle x2 + y2 – 2x – 4y – 4 = 0 which are
(i) perpendicular and(ii) parallel to the line 3x – 4y – 7 = 0.[Ans. (i) 4x + 3y = 25, 4x + 3y + 5 = 0, (ii) 3x – 4y + 20 = 0, 3x – 4y = 10 ]
Ex-2 Find the equation of the tangent drawn to the circle x2 + y2 – 6x + 4y – 3 = 0 from thepoint (7, 4) lying outside the circle. Also find the point of contact.
[Sol. y – 4 = m (x – 7) mx – y – 7m + 4 = 0
use p = r
2m1
4m72m3
= 4 2m1
6m4
= 4
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(2m – 3)2 = 4(1 + m2) 4m2 + 9 – 12m = 4 + 4m2
If m is finite 12
5 = m
If m is infinite then we cannot cancel 9m2. m . T : x = 7
and y – 4 =12
5(x – 7)
Ex-3 Find the equation of the tangent to the circle x2 + y2 = 4 drawn from the point (2, 3).[Sol. Method-I : y – 3 = m(x – 2)
use p = r
m ; m = 125Method-II : Equation of 'T' drawn to the circle x2 + y2 = 4 is
y = mx ± 2m12 passes from (2, 3) ; (3 – 2m)2 = 4 (1 + m)2
Note: With m = 125 we get 2 tangents which are parallel to each other and one ofthem passes through (2, 3).
Ex-4 Prove that the line y = x + 2 c touches the circle x2 + y2 = c2 and find its point ofcontact.
Ex-5 Tangent is drawn from the point P (4, 0) to the circle x2 + y2 = 8 touches it at the pointA in the 1st quadrant. Find the coordinates of another point B on the circle such thatAB = 4.
[Sol. cos =4
22 =
2
1
= 45° A (2, 2)Let B (x1, y1)Given AB = 4 (x1 – 2)2 + (y1 – 2)2 = 16
1121
21 y4x4yx = 8
Also 21
21 yx = 8
x1 + y1 = 0
21x2 = 8
x1 = ± 2 B (2, –2) or (–2, 2) ]
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Note: Point of intersection of the tangent drawn to the circle x2 + y2 = a2 at the pointP () and Q () is
h =
2cos
2cosa
; k =
2cos
2sina
EXAMPLES :Ex-1 Find the locus of the point of intersection of the pair of tangents drawn to a circle
x2 + y2 = a2 at P () and Q (), where | – | = 120°.
[Sol. h =
2cos
2cosa
; k =
2cos
2sina
h = 2a2
cos
& k = 2a2
sin
x2 + y2 = 4a2 ]
Ex-2 Equation of a straight line joining two points and on the x2 + y2 = a2 is
x2
cos
+ y2
sin
= a cos2
[Proof: Equation of line joining and is x cos + y sin = p
where = +2
=
2
and p = a cos2
(from figure)]
Ex-3 Prove that the perpendicular from any point on the circle x2 + y2 = a2 on the line joiningthe points of contact of 2 tangents is a mean proportional between the perpendicularsfrom the point upon the 2 tangents.
[Sol. TPT p2 = p1p2Equation of chord joining and
x2
cos
+ y2
sin
= a cos2
p =2
cos2
sinsin2
cos(cosa
p =
2
cos2
cosa
p = 2a ·
2sin
2sin ....(1)
Tangent at '' and ''x cos + y sin = a and x cos + y sin = a
p1 = | a (cos · cos + sin · sin – 1) | = a {cos ( – ) – 1}|||ly p2 = a {cos ( – ) – 1} p1 p2 = a2 {1 – cos ( – )}{1 – cos ( – )}
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= a2
2sin2
2sin2 22
p1 p2 = 4a2 sin2
2
sin2
2
....(2)
clearly p2 = p1 p2 Hence proved. ]
Length of Tangent & power of a point."Length of the tangent from an external point P(x1, y1) to a given circle:"
S = x2 + y2 + 2gx + 2fy + c = 0L2 + R2 = d2
L2 = d2– R2 = (x1 + g)2 + (y1 + f)2 – (g2 + f2 – c)
L2 = 21
21 yx + 2gx1 + 2fy1 + c
L1 = cfy2gx2yx 1121
21 = 1S
Note: All these formulae are applicable when coeff. of x2 & y2 is unity(i) Area of Quad PAOB = 2 POA
= 2 ·2
1R L = R L
(ii) Find AB i.e length of chord of contact
AB = 2 L sin where tan =L
R
= 22 LR
LR2
(iii) Area of PAB ( formed by pair of Tangent & corresponding C.O.C.)
PAB =2
1AB × PD
=2
1( 2 L sin) (L cos)
= L2 sin cos
= 22
3
LR
LR
(iv) Angle 2 between the pair of Tangents
tan 2 =2tan1
tan2 =
)RL(L
LR222
2
2 = tan–1
22 RL
RL2
(v) Equation of the circle circumscribing the PAB. ( One such circle described on OP asdiameter)i.e. (x – x1) (x + g) + (y – y1) (y + f) = 0
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Power of a point :Square of the length of the tangent from the point P is called power of the point P w.r.ta given circle i.e. PT2 = S1Note: Power of a point remains constant w.r.t a circle
PA · PB = (PT)2
Analytical proof :
cos
xx 1 =
sin
yy 1 = r ; substituting x = x1 + r cos and y = y1 + r sin in x2 + y2 = a2,
we get,
r2 + 2r (x1 cos + y1 sin) + 21
21 yx – a2 = 0
r1r2 =21
21 yx – a2 = constant = (PT)2
Note: Power of a point is + ve / 0 (zero) / – ve according as point 'P' liesoutside / on / inside the circle.
EXAMPLES :Ex-1 Find the length of the Tangent from any point on the circle
S1 x2 + y2 + 2g1x + 2f1y + = 0 to the circle
S2 x2 + y2 + 2gx + 2fy + = 0 is
[Sol. P(x1y1) on S1 2S
= λμ ]
Ex-2 Find the value of 'p' for which the power of a point P(2, 5) is negative w.r.t a circlex2 + y2 – 8x – 12y + p = 0 and the circle neither touches nor intersects the coordinate axis.
[Sol. 4 + 25 – 16 – 60 + b < 0p < 47 ....(1)also g2 < c & f2 < c16 < p ....(2)36 < p ....(3)from (1), (2) and (3) 36 < p < 47 ]
Ex-3(a) A point moves such that a tangent from it to the circle x2 + y2 + 4x – 5y + 6 = 0 is
double the length of the tangent to the circle x2 + y2 = 4. Show that the locus is a circle.
Find its centre and radius. [Ans. centre :
6
5,
3
2 ; radius =
6
305]
[Sol. Let the point be P (x1, y1) and let the lengths of the tangents from P to the two circles bet1 and t2 respectively.
Then 6y5x4yxt 1121
21
21 and 4yxt 2
121
22
but t1 = 2t2 22
21 t4t
6y5x4yx 1121
21 = )4yx(4 2
121
i.e. 022y5x4y3x3 1121
21
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Dividing throughout by 3, we get the equation as
x2 + y2 –3
4x +
3
5y –
3
22 = 0
Its centre is
6
5,
3
2 and radius =
6
305]
(b) Find the locus of a point the tangents from which to the circles 4x2 + 4y2 – 9 = 0 and9x2 + 9y2 – 16 = 0 are in the ratio 3 : 4. [Ans. 7x2 + 7y2 = 20]
H.W. after Lecture -3 : Exercise - 18 (Loney) Except Q.No. 21
LECTURE – 48. DIRECTOR CIRCLE:
Director circle is a name given to a special locus. Locus of a point 'P' which moves insuch a way such that the pair of tangents drawn from 'P' to a given curve makes anangle of 90°.
ORLocus of the point of intersection of two mutually perpendicular tangents drawn to agiven curve is called the director circle of the given curve.
(Director circle of a parabola is its own directrix)
Method-I : y = mx ± 2m1a passes through (h, k) (k – mh)2 = a2 (1 + m2) m2(h2 – a2) – 2mhk + k2 – a2 = 0
m1m2 = –1 22
22
ah
ak
= –1
x2 + y2 = 2a2
Director circle of a circle is a concentric circle having radius equals to 2 timesif equation of circle is x2 + y2 +2gx + 2fy + c = 0 then its director circle is
(x + g)2 + (y + f)2 = 2R2
R = 222 cfg Method-II : OAPB is square
a2 + a2 = OP2
2a2 = h2 + k2
x2 + y2 = 2a2
EXAMPLES :Ex-1 Locus of point 'P' which moves such that the angle made by the pair of tangents drawn
to the circle x2 + y2 = a2 is 60°.
[Sol. sin 30° = 22 kh
a
i.e. h2 + k2 = 4a2 or x2 + y2 = 4a2.
Alternate : use tan260° = 221
221
)mm1(
)mm(
3 = 221
212
21
)mm1(
mm4)mm(
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Ex-2 Find the range of values of 'a' such that the angles '' between the pair of tangents
drawn from the point (a, 0) to the circle x2 + y2 = 1 satisfies2
< < .
[Sol.Draw the director circle of x2 + y2 = 1i.e. x2 + y2 = 2
a ( 2 , – 1) (1, 2 ) Ans. ]
Chord of a circle with a given middle point M (x1, y1)
y – y1 = –1
1
y
x(x – x1)
y y1 –21y = 2
1x – x x1
x x1 + y y1 =21
21 yx
x x1 + y y1 – a2 = 21
21 yx – a2
T = S1 (not to be used in case of circle) ....(1)
Note: Of all the chords which passes through a given point M (a, b) inside the circlethe shortest chord is one whose middle point is (a, b)
Proof : From the figure
l = 22 pR ....(1)
and L = 22 qR ....(2)
clearly p > q l < L 2l < 2L Hence proved.
LAB = 2 22 pR = 2 })fb()ga{()cfg( 2222
LAB = 2)cbf2ag2ba(
negative
22
(since a, b lies inside the circle)
EXAMPLES :Ex-1 Locus of the middle point of the chords of the circles x2 + y2 +2gx + 2fy + c = 0 which
passes through a fixed point (a, b) lying outside the circle.[Sol. m1 m2 = – 1
ah
bk·
gh
fk = – 1
(x – a) (x + g) + (y – b) (y + f ) = 0Locus represent the arc of a circle described on CP as diameter. ]
Ex-2 Find the equation to the locus of the middle point of the chord of the circlex2 + y2 + 2gx + 2fy + c = 0 which subtends right angle at a given point P (a, b)
[Sol. From figureAC2 = AM2 + CM2
but AM = PM AC2 = PM2 + CM2
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Ex-3 Find the equation to the locus of the feet of the perpendicular drawn from the originupon a variable chord of the circle x2 + y2 + 2gx + 2fy + c = 0 which subtends rightangle at origin.
[Sol. Equation of chord AB with mid point M (h, k)
y – k = –k
h (x – h)
ky – k2 = h2 – hx hx + ky = h2 + k2 ....(1)Now homogenising S = 0 with the help of (1)
x2 + y2 + 2 (gx + fy)
22 kh
kyhx + c
2
22 kh
kyhx
= 0
Coefficient of x2 + coefficient of y2 = 0 ]
9. CHORD OF CONTACT :Definition : It is defined as the straight linejoining the point of contact of the pair oftangents drawn from an external point P (x1, y1)to a circle.
Note: Chord of contact will exist only if the point P does not lie inside the circle.Equation of chord of contact AB :Tangent at 'A' x x' + y y' = a2
passes through (x1, y1) x1 x' + y1 y' = a2 ....(1)|||ly at 'B' x x" + y y" = a2
passes through (x1, y1) x1 x" + y1 y" = a2 ....(2)from (1) and (2) equation if chord of contact can be written by replacing x', x" by xand y', y" by y.hence, x x1 + y y1 = a2 ....(3) (same as that of tangent)
(i) Equation (3) is a linear relation in x and y it represent a line.
(ii) Point 'A' and 'B' satisfy equation (3) it represent the equation of chord of contact. For general equation of circlechord of contact xx1 + yy1 + g(x – x1) + f (y – y1) + c = 0
EXAMPLES :Ex-1 Tangents are drawn to the circle x2 + y2 = 8 at the points where the line x + 2y = 4
intersect the circle. Find the coordinates of the point of intersection of the tangents.[Sol. Equation of C.O.C. AB is
x x1 + y y1 = 8 ....(1)Also equation of AB is
x + 2y = 4 ....(2)(1) and (2) must be identical
1
x1 =2
y1 =4
8 x1 = 2 ; y1= 4 ]
Ex-2 Tangents are drawn to a unit circle with centre at origin from every point on the line2x + y = 4, prove that(i) chord of contact passes through a fixed point(ii) equation to the locus of the middle point of chord of contact.
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[Sol.(i) 'P' is a variable point moving on line
equation of chord of contact isx x1 + y y1 = 8 ....(1)
also 2x1 + y1 = 4 ....(2) x x1 + y (4 – 2x1) = 1
x1 (x – 2y) + (4y – 1) = 0
it represent a family of line passes through a fixed point
4
1,
2
1.
(ii) M (h , k)equation of chord with variable given middle point (h, k)is xh + yk = h2 + k2 ....(1)Also C.O.C is x x1 + y y1 = 1 ....(2)
comparing (1) and (2)h
x1 =k
y1 = 22 kh
1
x1 = 22 kh
h
; y1 = 22 kh
k
Also 2x1 + y1 = 4 2h + k = 4 (h2 + k2) 2x + y = 4 (x2 + y2) ]
Ex-3 Chord of contact of the tangent drawn from a point on the circle x2 + y2 = a2 to thecircle x2 + y2 = b2 touches the circle x2 + y2 = c2. Prove that a, b, c are in G.P.(a, b, c > 0)
[Sol. Equation of COC AB is x x1 + y y1 = b2
touches the circle x2 + y2 = c2 c =21
21
2
yx
b
Also 21
21 yx = a2
c = a
b2
b2 = ac hence proved. ]
Ex-4 If 2 distinct chords of the circle x2 + y2 – ax – by = 0 drawn from the point (a, b) isdivided by the x-axis in the ratio 2 : 1 then prove that a2 > 3b2.
[Sol. We have, 0 =3
k2b·1 k = –
2
b
hence y = –2
b satisfy the given circle x2 + y2 – ax – by = 0
x2 +4
b2 – ax + b ·
2
b = 0
x2 – ax +4
b3 2 = 0. For two distinct values of h,
D > 0 a2 – 3b2 > 0 a2 > 3b2 ]
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Ex-5 If the chord of contact of tangents drawn from P to the circle x2 + y2 = a2 subtends aright angle at the centre, find the locus of P.
[Sol. Let P be the point (x1, y1)Then QR is x x1 + y y1 = a2
The combined equation CQ and CR is
x2 + y2 = a2
2
211
a
yyxx
i.e. x2
2
21
a
x1 – 2
11
a
yxyx2 + y2
2
21
a
y1 = 0
Since QCR = 90°, coefficient of x2 + coefficient of y2 = 0
2
21
a
x1 + 1 – 2
21
a
y1 = 0
i.e. 221
21 a2yx
The locus of (x1, y1) is the concentric circle x2 + y2 = 2a2
Alternate : PQCR is a square 221
21 a2yx ]
10. PAIR OF TANGENTS :Combined equation of the pair of tangents drawn from an external point 'P' to a given
circle is SS1 = T2 , where S x2 + y2 – a2 ; S1 22
121 ayx .
Proof : Equation of COC AB is x x1 + y y1 = a2
triangles AQN and APM are similar
AP
AQ =
PM
QN
L
l =
2
1
p
p
221
21
222
ayx
akh
= 22
121
211
ayx
akyhx
(x2 + y2 – a2) ( 21
21 yx – a2) = (x x1 + y y1 – a2)2
S S1 = T2 ]EXAMPLES :Ex-1 Show that the equation to the pair of tangents drawn from the origin to the circle
x2 + y2 + 2gx + 2fy + c = 0 is (gx + fy)2 = c(x2 + y2)
Ex-2101/05 Tangents are drawn to the circle x2 + y2 = a2 from two points on the axis of x, equidistantfrom the point (k, 0). Show that the locus of their intersection is ky2 = a2(k – x)
[Sol. SS1 = T2
( 221
21 ayx ) [(k – )2 – a2] = [(k – )x1 – a2 ]2 ....(1)
again( 22
121 ayx ) [(k + )2 – a2] = [(k + )x1 – a2 ]2 ....(2)
(2) – (1) gives4( 22
121 ayx ) k = [(k + ) x1 – a2 + (k – ) – a2] [(k + )x1 – (k – )x1]
= [2 (kx1 – a2)] [2x1] = 4x1 [kx1 – a2] = k ( 22
121 ayx ) = k 2
1x – a2x1
= k 21y = a2(k – x1)
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LECTURE - 511. FAMILY OF CIRCLES :
In how many different models we can lay down the equation of family of circles.Type-1 : Equation of the family of circles which passes through the points of intersection of
two circles S1 = 0 and S2 = 0 isS1 + S2 = 0 – 1 (Justify the family)
Find the equation of a circle which passes through the point of intersection of S1= 0and S2 = 0S1 : x
2 + y2 – 4x + 6y – 3 = 0S2 : x
2 + y2 + 4x – 6y + 12 = 0(i) which passes through (0, 0) / (a, b)
(ii) Radius = 5 ; cfg 22 = 5
(iii) x-intercept = 5 i.e. 2 cg2 = 5
(iv) y-intercept = 5 i.e. 2 cf 2 = 5
(v) Line 3x + 4y – 5 = 0 is tangent to the circle i.e. p = r(vi) Centre lies on x-axis i.e. f = 0
or centre lies on y-axis i.e. g = 0
(vii) line 4x – 7y = 10 is normal to the circle i.e. centre lies on the line.[Hint: Equation of the family of circle is S1 + S2
i.e. x2 (1 + ) + y2(1 + ) – 4x (– ) + 6y (1 – ) + 12 – 3 = 0 ]
Type-2 : Equation of the family of circles passes through the pointof intersection of a circle S = 0 and a line L = 0 is given by
S + L = 0Conversely the eqution S + L = 0 represents a family of circles passing through twofixed points which the points of intersection of L = 0 and S = 0.
EXAMPLES :Ex-1 Find the equation of a circle drawn on the chord x cos + y sin = p of the circle
x2 + y2 = a2 as its diameter.[Sol. Equation of family of circles is (x2 + y2 – a2) + (x cos + y sin – p) = 0
Now centre lies on the line x cos + y sin = p
i.e. –2
cos2 –
2
sin2 = p
= – 2p ]
Ex-2 Show that the equation x2 + y2 – 2x – 2y – 8 = 0 represents for different values of ,a system of circles passing through two fixed points A and B on the x-axis, and alsofind the equation of that circle of the system the tangent to which at A and B meet onthe line x + 2y + 5 = 0.
[Sol. (x2 + y2 – 2x – 8) – 2·y = 0S + L = 0Solving S = 0 & L = 0put y = 0 x2 – 2x – 8 = 0
(x – 4) (x + 2) = 0x = – 2 or 4
A (4, 0) B (–2, 0)
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Equation of AB :C.O.C. xx1 + yy1 –1 (x+x1) – (y + y1) – 8 = 0 x(x1 –1) + y (y – y1) – (x1 + y1+ 8) = 0 ....(1)Also equation of AB is x-axis i.e
0·x + 1·y + 0·c = 0 ....(2)comparing (1) & (2)
1x
0
1 =
1y
1 =
8yx
0
11
x1 = 1 & y1+ 9 = 0
y1 = 9
Also x1 + 2y1 + 5 = 0 1 –18
+ 5 = 0 =18
= 3
equation of circle is x2 + y2 – 2x – 6y – 8 = 0 ]
Ex-3 Find the equation of a circle which passes through origin and through the point ofcontact of the tangents drawn from the origin to the circle x2 + y2 – 11x + 13y + 17 = 0
[Sol. Equation of COC AB is
x·0 + y·0 –2
11(x + 0) +
2
13( y + 0) + 17 = 0
i.e. 11x – 13y – 34 = 0Now equation of family of circle isx2 + y2 – 11x + 13y +17 + (11x – 13y – 34) = 0passes through (0, 0) ]
Type-3: Equation of the family of circles passes through two given points A (x1, y1) & B(x2, y2) is:
(x–x1) (x–x2) + (y–y1) (y–y2) + 1yx1yx1yx
22
11 = 0
Note: Fixed circle of this family is the circle described onAB as diameter.
Ex-4 Two circles are drawn through the point (a, 5a) & (4a, a) to touch the axis of 'y'. Prove
that they intersect at an angle of tan–19
40
[Sol. Family of circle's passes through two fixed points is given by:(x–a) (x–4a) + (y–5a) (y–a) + L = 0L 4x + 3y = 19ax2 + y2 – 5ax – 6ay + 4a2 + 5a2 + ( 4x + 13y – 19a) = 0touches y-axis f2 = L & now proceed ]
Type-4 : Equation of family of circles touching a line at its fixed point (x1, y1) is:(x–x1)
2 + (y–y1)2 + L = 0
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EXAMPLES:Ex-1 Find the equation of a circle which touches the line 2x–y = 4
at the point (1, –2) and(a) Passes through (3, 4)(b) Radius = 5
Ex-2 Find the equations of the circle which passes through the point (–1, 2) & touches thecircle x2 + y2 – 8x + 6y = 0 at origin.
[Sol. Equations of tangent is: 4x – 3y = 0 equation of the family of the circle is(x – 0)2 + (y – 0)2 + (4x – 3y) = 0passes through P (–1, 2) & get
Note:(i) Equation of a circle circumscribing a triangle whose sides are given by l1 = 0;
l2 =0 & l3 = 0 is given by l1l2 + l2l3 + l3l1 = 0 ....(1)provided coefficient of x2 = coefficient of y2 & coefficient of xy = 0(a) Represent a 2nd degree curve & A, B, C satisfy equation (1).This could also be used to compute circumcentre of a trianglewithout finding the coordinates of its vertices.
(ii) Equation of a circle circumscribing a quadrilateral whose sides inorder are represented by the lines l1 = 0; l2 =0 ; l3 = 0; l4 = 0 is givenby l1l3 + l2l4 = 0 ....(1)(a) represents a second degree curve & also A, B, C, D satisfy ....(1) l1l3 + l2l4 = 0 is the equation of circle provided coefficient ofx2 = coefficient of y2 & coefficient of xy = 0
LECTURE - 6
12. POLE & POLAR :(i) If through a point P in the plane of the circle , there be drawn any straight line to
meet the circle in Q and R, the locus of the point of intersection of the tangentsat Q & R is called the Polar of the point P and P is called the Pole of the Polar.
(ii) The equation to the polar of a point P (x1 , y1) w.r.t. the circle x2 + y2 = a2 is given byxx1 + yy1 = a2, & if the circle is general then the equation of the polar becomesxx1 + yy1 + g (x + x1) + f (y + y1) + c = 0. Note that if the point (x1, y1) be on the circlethen the chord of contact, tangent & polar will be represented by the same equation.
(iii) Pole of a given line Ax + By + C = 0 w.r.t. any circle x2 + y2 = a2 is
C
aB,
C
aA 22
.
(iv) If the polar of a point P pass through a point Q, then the polar of Q passes through P.
Explanation :
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Equation of the polar:Equation of the variable chord of contact of the point (h, k) w.r.t x2 + y2 = a2 is:
xh + yk = a2
passes through x1, y1 x1h + y1k = a2
required equation is xx1 + yy1 = a2
in case if circle is general then equation of polar becomesx x1 + y y1 + g (x + x1) + f (y + y1) + c = 0 & (x1, y1) is the pole.
EXAMPLE :Ex-1 To find the pole of a given line lx + my = n w.r.t a circle x2 + y2 = a2
i.e we have to find the coordinates of the point whose polar w.r.t the given circle is theline lx + my = nLet the pole be P (x1, y1) its polar w.r.t. the circle x2 + y2 = a2 is
x x1 + y y1 = a2 ....(1)also l x + m y = n ....(2)
(1) and (2) must be identical
n
a
m
y
n
xa 211
2
l
x1 = n
a2l and y1 = n
ma2
(not to be remember)
13. SOME DEFINITIONS :(i) Conjugate Points : Two points P and Q are said to be the conjugate of each other
w.r.t. the circle if polar of 'p' passes through 'Q' and vice versa.
(ii) Conjugate Lines: Two lines l1 and l2 are conjugate of each other if pole of one lieson other and vice versa.
EXAMPLES :Ex-1 find the value of 'k' for which the points (2, k) and (k, 3) are conjugate of each other
w.r.t. the circle x2 + y2 = 10.[Sol. Polar of (2, k) is 2x + ky = 10 ....(1)
passes through (k, 3)2k + 3k = 10 k = 2 ]
Ex-2(a) The lengths of the tangents from two points A and B to a circle are l1 and l2 respectively.If the points are conjugate with respect to the circle x2 + y2 = a2 then, show that(AB)2 = l1
2 + l22.
[Hint: Let A (x1, y1) and B (x2, y2).
Now 21
221
21 ayx l and 2
222
222 ayx l .
Also x x1 + y y1 = a2 contains (x2, y2)Hence x1 x2 + y1 y2 = a2. Now (AB)2 = (x1 – x2)
2 + (y1 – y2)2 result ]
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(b) Prove that if two lines at right angles are conjugate with respect to a circle, one of themmust pass through the centre.
[Hint: Let the line be ax + by + c = 0 ....(1)and bx – ay + d = 0 ....(2)now pole of (1) passes through (2) d = 0 ]
Ex-3 A variable circle is drawn to touch the axis of 'x' at origin. Find the locus of the pole ofthe straight line lx + my + n = 0 w.r.t. the circle.
[Sol. Equation of the variable circle is(x – 0)2 + (y – )2 = 2
x2 + y2 – 2y = 0 ....(1)Let the pole of lx + my + n = 0 be (h, k) its polar w.r.t. (1) is
hx + ky – (y + k) = 0i.e. hx + y (k – ) – k = 0 ....(2)Also polar is lx + my + n = 0 ....(3)
on comparingl
h =
m
k = –
n
k
k – =l
mh and = lk
hn ; k – lk
hn =
l
mh or k2l – nh = mkh
required locus is ly2 = nx + mxy ]
Alternative Definition of polar :Locus of the harmonic conjugate of a given point P (x1, y1) w.r.t. 2 given points inwhich any line through P cuts a given circle x2 + y2 = a2 is polar of 'p' w.r.t. the givencircle i.e. x x1 + y y1 = a2.
x1 = 1
cosacosa
y1 = 1
sinasina
h =1
cosacosa
k =1
sinasina
h x1 + k y1 = 1
axa2
222
= a2
x x1 + y y1 = a2 Hence proved. ]
COMMON TANGENTS TO TWO CIRCLES
(1) Direct Common Tangent (DCT) (2) Transverse Common Tangent (TCT) (External common tangent) (Internal common tangent)
How do we distinguished between D.C.T. and T.C.T.
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Direct Common Tangent (D.C.T) : Here the centres of both the circles lie on thesame side of the tangent line.
Transverse Common Tangent (T.C.T.) : Here the centres of both the circles lie on theopposite side of the tangent line.
POSITION OF THE CIRCLES (AND NUMBER OF COMMON TANGENTS) :(1) If 2 circles are separated, then d > r1 + r2 and in this case there exists
4 common tangent.T.C.T2
.T.C.D2
(2) If 2 circles touche externally then d = r1 + r2 and in this case there exist
3 common tangents
)contactofintpotheirat(.T.C.T1
.T.C.D2
(3) If 2 circles intersect each other then | r2 – r1 | < d < r1 + r2and in this case there exist2 common tangent both of them being D.C.T.
Note: Only d < r1 + r2 is not the sufficient condition inthis case because if one circle is contained in otherthen also d < r1 + r2.
EXAMPLES :Ex-1 Find the range of 'r' so that the circles :
(x – 1)2 + (y – 3)2 = r2
and (x – 4)2 + (y + 1)2 = 9intersects at 2 distinct points ('r' being the radius of the circle.)
[Sol. Method-I : for circles to intersectingr > 5 – 3
and r < 8 range of 'r' 2 < r < 8
Method-II : d = 5 for circle to intersect
| r1 – r2 | < d < r1 + r2 | r – 3 | < 5 < r + 3 r > 2 ....(1)and | r – 3 | < 5 – 5 < (r – 3) < 5
– 2 < r < 8 ....(2) 2 < r < 8 ]
(4) If both the circles touches each other internally, then d = | r1 – r2 |and in this case there exist only one common tangent i.e. D.C.T. attheir point of contact.
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(5) If one circle is contained in other then d < | r1 – r2 | and in this casethere exist no common tangent.
LENGTH AND EQUATION OF COMMON TANGENTS :To find analytically the length of common tangent (both external and internal commontangent)
From figure : 2extL + (r1 – r2)
2 = d2
Lext =2
212 )rr(d
d2 = 2intL + (r1 + r2)
2
Lint =2
212 )rr(d
How to find equation of D.C.T / T.C.T.Statement :(a) Direct Common Tangent (D.C.T.) meet at a point which divides the line joining the
centre of circles externally in the ratio of their radii.(b) Transverse Common Tangent (T.C.T.) meets at a point which divides the line joining
the centres of circles internally in the ratio of their radii.
Proof :2
1
2
1
r
r
PC
PC
(since for S2 = 0, C2P is the angle bisectorand also for S1 = 0, C1P is the anglebisector. C1, C2, P will lie in a same line
|||ly2
1
2
1
r
r
QC
QC
Note: C1 ; Q, C2, P constitute a harmonic rangei.e. C1Q, C1C2, C1P are in H.P.
EXAMPLES :Ex-1 Equation of transverse common tangent is x = 0 and y = 0
from figure
Ex-2 If the figure isone of the equation of D.C.T. is x = 0 and T.C.T. is y = 0
Ex-3 Prove that the common tangents to the circles x2 + y2 – 6x = 0 and x2 + y2 + 2x = 0from an equilateral triangle.
[Sol. S1 : x2 + y2 – 6x = 0 C1 (3, 0) ; r1 = 3
S2 : x2 + y2 + 2x = 0 C2 (– 1, 0) ; r2 = 1
d = 4 d = r1 + r2
circles touches externally.
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sin =2PC
1
and sin =1PC
3
eccos2PCPC 21
C1 C2 = 2 cosec 4 = 2 cosec
sin =2
1 = 3 2 = 60°
ABP is equilateral.
Ex-4 Find all the common tangent to the circles x2 + y2 = 1 and (x – 1)2 + (y – 3)2 = 4.(note that circles are separated)
Ex-5 Find the equation of the circles to which the line 4x + 3y = 10 isa common tangent at (1, 2) and radius of each of the circle is 5.
[Sol. Method-I : Family of circles touches the line at (1, 2) IS(x – 1)2 + (y – 2)2 + (4x + 3y – 10) = 0
now r = 5Method-II : From (1, 2) use parametric to get the coordinates of C1 and C2. ]
14. RADICAL AXIS AND RADICAL CENTRE :Radical axis of 2 circles is the locus of a point whose powers w.r.t. the two circles areequal. The equation of radical axis of two circles S1 = 0 and S2= 0 is given by
S1 – S2 = 0 i.e. 2 (g1 – g2) x + 2 (f1 – f2) y + c1 – c2 = 0
h2 + k2 + 2g1h + 2f1k + c1 = h2 + k2 + 2g2h + 2f2k + c2
2(g1 – g2) x + 2 (f1 – f2) + c1 – c2 = 0is the required equation of radical axis of S1 = 0 and S2 = 0
Note that(a) If two circles intersect, then the radical axis is the common
chord of the two circles.
(b) If two circles touch each other then theradical axis is the common tangent of thetwo circles at the common point of contact.
(c) Radical axis is always perpendicular to the linejoining the centres of the two circles.
(d) Radical axis need not always pass through the mid point of the line joining the centresof the two circles.
(e) Radical axis bisects a common tangent between the two circles.
(f) Pairs of circle which do not have radical axis are concentric.S1 = x2 + y2 + 2gx + 2fy + c1 = 0S2 = x2 + y2 + 2gx + 2fy + c2 = 0
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EXAMPLES :Ex-1 Show that the equation of a straight line meeting the circle x2 + y2 = a2 in 2 points at
equal distance 'd' from the point (x1, y1) on its circumference is :
x x1 + y y1 – a2 +2
d2
= 0
[Sol. Equation of a circle with centre 'p' and radius 'd'i.e. (x – x1)
2 + (y – y1)2 = d2
or x2 + y2 – 2xx1 – 2yy1 +21
21 yx = d2 ; also 22
121 ayx
the required line is the radical axis between the circlesx2 + y2 – 2xx1 – 2yy1 + a2 – d2 = 0 ....(1)
and x2 + y2 – a2 = 0 ....(2) equation of AB is (1) – (2)
– 2 x x1 – 2 y y1 + 2a2 – d2 = 0
i.e. x x1 + y y1 – a2 +2
d2
= 0 Hence proved ]
Ex-2 Prove that the circle x2 + y2 + 2gx + 2fy + c = 0 will bisect the circumference of thecircle x2 + y2 + 2g'x + 2f'y + c' = 0 if 2g' (g – g') + 2f' (f – f') = c – c'.
[Sol. S1 : x2 + y2 + 2gx + 2fy + c = 0
S2 : x2 + y2 + 2g'x + 2f ' y + c' = 0
If S1 bisects the circumference of S2 then their common chord is the diameter ofS2 = 0.equation of common chord
S1 – S2 = 02 (g – g') x + 2(f – f')y + c – c' = 0 ....(1)
(–g, – f') lies on equation (1)– 2g' (g – g') – 2 f '(f – f ' ) + c – c' = 0
2g' (g – g') + 2 f '(f – f ' ) = c – c' hence proved. ]
Ex-3 Tangent are drawn to the circle x2 + y2 = 12 at the points where it is met by the circlex2 + y2 – 5x + 3y – 2 = 0. Find the point of intersection of the tangents.
[Sol. equation of AB :(x2 + y2 – 5x + 3y – 2) – (x2 + y2 – 12) = 05x – 3y – 10 = 0 ....(1)
Also AB is chord of contact w.r.t the circle x2 + y2 = 12i.e. x x1 + y y1 – 12 = 0 ....(2)comparing (1) and (2)
10
12
3
y
5
x 11
x1 = 6 & y1 = –5
18Ans. ]
Ex-4 Consider a family of circles passing through two fixed points A (3, 7) and B (6, 5).Show that the chords in which the circles x2 + y2 – 4x – 6y – 3 = 0 cuts the member ofthe family are concurrent at a point. Also find the coordinates of this point.
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[Sol. Family of circles passes through two fixed point isS + L = 0 ....(1)
where S = (x – 3) (x – 6) + (y – 7) (y – 5) = 0
L =1561731yx
= 0
Equation of variable common chord isS + L – S' = 0
(S – S') + L = 0or L' + L = 0 which represents family of lines concurrent at L' = 0 and L = 0.
Ex-5 Prove that the circle x2 + y2 + 2ax + c2 = 0 and x2 + y2 + 2by + c2 = 0 touche each other
if 222 c
1
b
1
a
1
[Sol. Subtract to get common tangent and drop perpendicular from centre on any one circleand equate it to its radius.]
Ex-6 Find the equation of a circle which bisects the circumferences of the circles x2 + y2 = 1,x2 + y2 + 2x = 3 and x2 + y2 + 2y = 3. [Ans. x2 + y2 + 4x + 4y – 1 = 0]
Ex-7 Find the locus of the centre of circles which bisect the circumference of the circlesx2 + y2 = 4 and x2 + y2 – 2x + 6y + 1 = 0. [Ans. 2x – 6y – 15 = 0]
Ex-8 Given S = x2 + y2 + 2gx + 2fy + c = 0 and two point Aand B on it. Prove that the equation of the line AB is(x1 + x2 + 2g)x + (y1 + y2 + 2f)y = x1x2 + y1 y2 – c
Ex-9 Find the equation of the circle which bisects the circumference of the circlex2 + y2 + 2y – 3 = 0 and touches the line x – y = 0 at origin.
RADICAL CENTRE :Definition : The common point of intersection of the radical axis of 3 circles taken2 at a time is called the Radical Centre of three circles. from this
Radical Axis of S1 & S2 : 2 (g1 – g2)x + 2 (f1 – f2) y + c1 – c2 = 0Radical Axis of S2 & S3 : 2 (g2 – g3)x + 2 (f2 – f3) y + c2 – c3 = 0Radical Axis of S3 & S1 : 2 (g3 – g1)x + 2 (f3 – f1)y + c3 – c1 = 0
D =131313
323232
212121
ccffggccffggccffgg
= 0 ; Use R1 R1 + R2 + R3 D = 0
(solve any 2 radical axes to get radical centre which is a point from which tangent to thethree circles are equal.) www.StudySteps.inPage 28 of 32
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EXAMPLES :Ex-1 Lines lx + my + n = 0 ; px + qy + r = 0 intersects the circle
x2 + y2 + 2g1x + 2f1y + c1 = 0 and x2 + y2 + 2g2x + 2f2y + c2 = 0 at 4 points A and B ;C and D respectively. If these points are concyclic then prove that
rqpnm
cc)ff(2)gg(2 212121l
= 0
[Hint: The 3 radical axis are concurrentlx + my + n = 0px + qy + r = 0
and S1 – S2 = 0use condition of concurrency ]
Ex-2 Prove analytically that radical centre of 3 circlesdescribed on the sides of a triangle as diameter is theorthocenter of the triangle. (Remember this)
[Sol. (check slope of radical axes and BC) (m1m2 = –1) Radical axes will act as an altitude. ]
COAXIAL SYSTEM OF CIRCLES :Definition : A system of circles, every 2 of which have the same radical axis, is calledCoaxial system of circles.
or
EXAMPLES :Ex-1 Find the equation of a circle coaxial with S1 : x
2 + y2 + 4x + 2y + 1 = 0 andS2 : 2x2 + 2y2 – 2x – 4y – 3 = 0 and the centre of the circle to be determined lies onthe radical axes of these 2 circle.
[Sol. S1 : x2 + y2 + 4x + 2y + 1 = 0
S2 : x2 + y2 – x – 2y –
2
3 = 0
equation of radical axis S1 – S2 = 0
i.e. 5x + 4y +2
5 = 0
family of circles passes through S1 = 0 and L = 0S1 + L = 0 ....(1)
centre lies on L = 0, get .
Ex-2 Find the equation of the circle passes through (1, 1) belonging to the system of coaxialcircles which touches the locus of the point of intersection of 2 perpendiculars tangentsto the circle x2 + y2 = 4 at (2, 2).
Ex-3 From a point P tangents drawn to the cirlces x2 + y2 + x – 3 = 0 ;3x2 + 3y2 – 5x + 3y = 0 and 4x2 + 4y2 + 8x + 7y + 9 = 0 are of equal length. Find theequation of the circle passes through P and which touches the line x + y = 5 at (6, –1)
[Ans. x2 + y2 – 7x + 7y + 12 = 0]www.StudySteps.inPage 29 of 32
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ORTHOGONALITY OF TWO CIRCLES :Definition : Two curves are said to be orthogonal if they intersect each other at 90°wherever they intersects.
Condition for orthogonality of 2 circles :22
21 rr = (g1 – g2)
2 + (f2 – f2)2
(1
21
21 cfg ) + (
222
22 cfg ) = (g1 – g2)
2 + (f1 – f2)2
2 g1 g2 + 2 f1 f2 = c1 + c2 (condition for orthogonality)
EXAMPLES :Ex-1 If the circles, S1 : x
2 + y2 + 2x + 2ky + 6 = 0 and S2 : x2 + y2 + 2kx + k = 0 intersects
orthogonally then find k.
Ex-2 Find the equation of a circle which passes through the point of intersection of S1 = 0and S2 = 0 and cuts the circle S3 = 0 orthogonally.
Ex-3 Prove that locus of the centre of a variable circle x2 + y2 + 2gx + 2fy + c = 0 whichcuts the 2 given circles x2 + y2 + 2g1x + 2f1y + c1=0 and x2 + y2 + 2g2x + 2f2y + c=0orthogonally is the radical axis of 2 given circles.
[Sol. Let locus be (h, k)Given : 2 g g1 + 2 f f1 = c + c1 ....(1)
2 g g2 + 2 f f2 = c + c2 ....(2)(1) – (2) 2g (g1 – g2) + 2f (f1 – f2) = c1 – c2Replace g – h and f – k
2x (g1 – g2) + 2y (f1 – f2) = c2 – c1 radical axis ]
Note : If a line is orthogonal to a circle means the line is normal / diameter of the circle.
Ex-481/04 Find the equation of the circle which cuts the circle x2 + y2 + 2gx + 2fy + c = 0,the lines x = – g and y = – f orthogonally.
[Sol. Consider the required circle having the equation x2 + y2 + 2g x + 2f y + c = 0 (*)It orthogonally cuts the circle x2 + y2 + 2gx + 2fy + c = 0
Using condition of orthogonality , we have 2gg + 2f f = c + c c = 2gg + 2f f – c ....(i)Two curves are said to be orthogonal if the tangents to the curves, drawn at the pointof intersection of the curves intersect at a right angle.
Required circle with orthogonally cut the line x = – g. Therefore the points of intersectionwill be extremities of the diameter of the required circle and equation of the diameter isx = – g
x - coordinate of the centre = – g = – gSimilarly we can argue and conclude that y coordinate of the centre = – f = – f
Centre of required circle (– g, – f)From (i) we get c = 2g2 + 2f 2 – c
equation of required circle x2 + y2 + 2gx + 2fy + 2g2 + 2f 2 – c = 0 ]Note : Here we observe that the two circles (one which is given and one which we obtain are
concentric circles). They will intersect only if they have the same radii or both of themare point circle (having zero radius). So ! one may think! , if they do not intersect howcan they be orthogonal? The answer is they intersect in imaginary points.(Refer " Coordinate geometry by S.L. loney, page 163 , 3rd paragraph).
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Ex-5 Prove that the 2 circles which passes through the 2 points (0, a) and (0, – a) and touchethe straight line y = mx + d will cut orthogonally if d2 = a2 (2 + m2)
[Sol. Let the equation of circle bex2 + y2 + 2gx + 2fy + c = 0 ....(1)
g ; f ; c
Equation (1) passes through (0, a) and (0, – a) a2 + 2af + c = 0and a2 – 2af + c = 0
———————add 2a2 + 2c = 0 a2 = – csubstitue f = 0 f1 = f2 = 0Now equation (1) becomes
x2 + y2 + 2gx + c = 0 ....(2)line y = mx + d touches (2)
2m1
dmg
= cg2
(mg – d)2 = (g2 – c) (1 + m2)m2g2– 2mgd + d2 = g2 + m2g2 – c – m2c
g2 + 2mgd – d2 – c(1 + m2) = 0 ....(3)
Also c = – a2 c = c1 = c2 = – a2
using condition of orthogonality2g1g2 = 2c = – a2
g1g2 = – a2
from equation (3) g1g2 = – {d2 + c(1 + m2)} d2 + c(1 + m2) = a2
c = – a2 d2 – a2(1 + m2) = a2
d2 = a2 (1 + m2) Hence proved. ]
Ex-6 The circles x2 + y2 + 2g1x + 2f1y + c1 = 0 and x2 + y2 + 2g2x + 2f2y + c2 = 0intersect orthogonally at A and B. If C, D are the centres of these circles, show that theequation of the circle passing through A, B, C, D is
2 (x2 + y2) + 2(g1 + g2) x + 2 (f1 + f2) y + c1 + c2 = 0.[Sol. Since the circles intersect orthogonally hence the quad. ABCD is an cyclic quad.
required equation of the circle will be the circle on CD are diameter. (x +g1) (x + g2) + (y + f1) (y + f2) = 0 ....(1)Also condition of orthogonally gives
2 g1 g2 + 2 f1 f2 = c1 + c2 ....(2)
x2 + y2 + (g1 + g2) x + (f1 + f2) y +2
cc 21 = 0]
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CIRCLE ORTHOGONAL TO 3 GIVEN CIRCLES :
Method-I: Let the equation of circle be x2 + y2 + 2gx + 2fy + c= 0orthogonal to S1 = 0 ; S2 = 0 ; S3 = 0 get 3 relation in g, f, c.
Method-II : Circle orthogonal to 3 given circles is one such circlewhose centre is the radical centre of 3 given circles andradiusequals to length of tangent drawn from radical centreto any one of the circles.
[Solve any 2 radical axis to get radical centre and then compute r = LT]
Method-III : (It is a problem)
EXAMPLES :Ex-1 If 2 circles are orthogonal then the polar of any point 'P' on the first circle w.r.t. the second
circle always passes through the other end of the diameter of first circle through 'P'and
hence equation of the circle orthogonal to 3 given can be regarded as a locus of a point'P' whose polar w.r.t. 3 circles are concurrent.
[Sol. Part-I
Polar of the point 'P' w.r.t. the circle x2 + y2 + 2gx + 2fy + c = 0 is(a cos) x + (a sin) y + g (x + a cos ) + f (y + a sin ) + c = 0 ....(1)
(1) passes through (– a cos, – a sin )– a2 + c = 0
Also 2g · 0 + 2f · 0 = c – a2
c = a2 Hence proved.Part-II
S and S1 are orthogonalS and S2 are orthogonalS and S3 are orthogonal
Polar of P w.r.t. S1, S2 , S3 are concurrent at Q. xh + yk + g1(x + h) + f1(y + k) + c1 = 0|||ly xh + yk + g2(x + h) + f2(y + k) + c2 = 0and xh + yk + g3(x + h) + f3(y + k) + c3 = 0 required equation of circle is
33333
22222
11111
cyfxgfygxcyfxgfygxcyfxgfygx
= 0 ]
FINAL HOME WORK : Ex-19 + Ex-22 (leave Q.No. 9, 15, 18) + Ex-23 (leave Q.no. 6) + Ex–24 (Do only 12, 13, 14, 15) of S.L. Loney.
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