Solved Examples On Electrochemistry - askIITians...Find the equation of circle having the lines x 2 + 2xy + 3x + 6y = 0 as its normals and having size just sufficient to contain the

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Solved Examples

Example 1:

Find the equation of the circle circumscribing the triangle formed by the lines x + y

= 6, 2x + y = 4, x + 2y = 5.

Solution:

Method 1.

Consider the equation

(x + y – 6) (2x + y – 4) + λ1 (2x + y – 4) (x + 2y – 5) + λ2

(x + 2y – 5) (x + y – 6) = 0 …… (1)

This equation is satisfied by the points of intersection of any two of the given

three lines, i.e. it is satisfied by the vertices of the triangle formed by the given lines.

Result:

Now if (i) represents a circle then

(a) coefficient of x2 = coefficient of y2 and

(b) coefficient of xy = 0

And

3 + 5λ1 + 3λ2 = 0 ⇒ λ1 = – 6/5

Substituting these values in (i) and simplifying

We get

x2+ y2 – 17x – 19y + 50 = 0 (Ans.)

Which is the equation of the required circle.



Method 2.

Solve the lines in pair to find the vertices of the triangle and then obtain theequation

of the circle through these three points.

Example 2:

Find the locus of the point of intersection of perpendicular tangents to the circle x2 +

y2 = 4.

Solution:

Method 1.

y = mx + 2 √(1+m2 ) is tangent to x2 + y2 = 4 for all values of ‘m’.

It passes through (h, k) if

k = mh + 2 √(1+m2 )

⇒ (k – mh)2 = 4 (1 + m2)

⇒ m2 (h2 – 4) –2h km + (k2 – 4) = 0 …… (i)

The roots m1 and m2 of this quadratic equation are the slopes of PT and PT’. If PT

and PT’ are at right angle then m1m2 = –1.

From (i)

m1m2 = (k2-4)/(h2-4)



⇒ (k2-4)/(h2-4) = –1

⇒ h2 + k2 = 8

Locus of P(h, k) is x2 + y2 = 8

Method 2:

Equation of tangent at the point T(2 cos θ, 2 sin θ)

is 2 cos θ x + 2 sin θ y = 4

i.e. x cos θ + y sin θ = 2 …… (i)

Tangent at T’ would be perpendicular to the tangent at T

If ∠ TOT’ = 90o

i.e. ∠AOT’ = 90 + θ

Co-ordinates of T’ are (2 cos (90 + θ), 2 sin (90 + θ))

Equation of tangent at the point T’ is

– 2 sin θ x + 2 cos θ y = 4

or, – x sin θ + y cos θ = 2 …… (ii)

Think:

How to get the locus of point P(h, k)?

Caution:



Do not simply square and add (i) and (ii). Though we get the required result, but that it

not the right approach.

Well, lines (i) and (ii) both pass through the point P(h, k)

⇒ h cos θ + k sin θ = 2 and

– h sin θ + k cos θ = 2

Now we want to get a relation between h and k and also eliminate θ.

The best way to do this is to square and ad above two

Equations and we get h2 + k2 = 8

⇒ Locus of P(h, k) is x2 + y2 = 8

Method 3:

OP2 = OT2 + TP2 (∵ ∆OTP is a right angled triangle)

⇒ OP2 = OT2 + OT’2

⇒ OP2 = 2R2

⇒ h2 + k2 = 2(4) = 8

∴ Locus of P(h, k) is x2 + y2 = 8.

Note:

x2 + y2 = 8 is director circle of the circle x2 + y2 = 4

Example 3:

Find the condition that the line 3x + 44y – p = 0 is tangent to the circle x2 + y2– 4x –

6y + 9 = 0

Solution:

Radius of the given circle is 2 and centre is (2, 3). So for line 3x + 4x – p = 0 to be



tangent to the circle we have,

|(3.2+4.3-p)/√(32+42 )| = 2.

|18 – p| = 10

18 – p = 10 ⇒ p = 8

18 – p = –10 ⇒ p = 28

Hence the required condition is p = 8 or p = 28. (Ans.)

Example 4

A circular plot of land in the form of a unit circle is to be divided into two equal parts

by the arc of a circle whose centre is on the circumference of the circular plot. Show that

the radius of the circular arc is 2 cos θ, where θ is given by sin 2θ – 2θ cos 2θ = π/2.

Solution:

Let O be the centre of the given circular plot of radius 1 i.e. OA = OB = OC = 1 and A

be any point on its circumference. Again BDC be the arc of the circle with centre A and

dividing the given circle into two equal parts. Let r the radius of the new circle, then AB =

AC = AD = r.

Let ∠AOB = θ

Then ∠OBA = ∠OCA = θ, and ∠AOB = (π – 2θ).

Now area ABDCA must be = 1/2 area of unit circle = (π (1)2)/2 = π/2 …… (i)

r/sin(π -2θ ) =1/(sin π ) ⇒ r = 2 cos θ.

Required Area APBDCQA = Area of sector ABDCA + area of sector OCQAPB – 2 area of

OAB



⇒ π/2 = 1/2 r2 (2θ) + 1/2 (1)2 (2π – 4θ) – 2 × 1/2 (1)2 sin (π – 2θ)

sin 2θ – 2θ cos 2θ = π/2 (Ans.)

Example 5:

Find the equation of the circle which passes through the point (2, 0) and whose

centre is the limit of the point of intersection of lines 3x + 5y = 1, (2 + c)x + 5c2 y = 1.

Solution:

Solving, 3x + 5y = 1, (2 + c)x + 5c2y = 1

We get,

when c → 1

Pause:

We will study limits in detail in module 5.

Now, we want to find out the equation of the circle which passes through (2, 0) and

has its centre at (2/5), –1/25).

Equation of the circle is

(x-2/5)2+(y+1/25)2=(2/5-2)2+(1/25)2

This is the required equation of the circle.

Example 6:

Find the equation of circle having the lines x2 + 2xy + 3x + 6y = 0 as its normals

and having size just sufficient to contain the circle, x(x – 4) + y(y – 3) = 0



Solution:

The combined equation of two normal of the circle is given by

x2 + 2xy + 3x + 6y = 0

⇒ (x + 3)(x + 2y) = 0

⇒ x = –3, x = –2y

Recall:

A normal to a circle always passes through the centre of the circle. Now solving these,

we get the co-ordinates of the centre of the circle as (–3, 3/2); because the two normal

intersect at the centre of the circle

The required circle just contains the circle

x(x – 4) + y(y – 3) = 0

i.e. x2 + y2 – 4x – 3y = 0 …… (i)

Hence the required circle will touch the circle given by (1) internally.

Let r be the radius of the required circle. Now the two circles given by (1)

= √([22+(3/2)2 ] )=5/2 and centre = (2,3/2)

Now the required circle will touch the circle (i) internally, if

We have distance between the centre of the two circles = difference between their radii.

⇒ √((-3-2)2+(3/2-3/2)2 )=(r-5/2)

⇒ r = 15/2

Hence the equation of the required circle is given by

(x + 3)2 + (y-3/2)2=(15/2)2

x2 + y2 + 6x – 3y – 45 = 0 (Ans.)



Example 7:

A tangent is drawn to each of the circles x2 + y2 = a2, x2 + y2 = b2. Show that if these

two tangents are perpendicular to each other, the locus of their point of intersection is a

circle concentric with the given circles.

Solution:

Method 1:

Let P ≡ (x1, y1) be the point of intersection of the tangents PA and PB where A, B are

points of contact with the circles respectively.

As PA is perpendicular to PB, the corresponding radii OA and OB are also perpendicular.

Let ∠AOX = θ

⇒ ∠BOX = 90 + θ

Using the parametric from of the circle we can take

A ≡ (a cos θ, a sin θ)

∠ B ≡ (–b sin θ, b cos θ)

The equation of PA is

x (a cosθ) + y (a sin θ) = a2

y cos θ – x sin θ = b

Since P(x1, y1) lies on these tangents

⇒ x1 cos θ + y1 sinθ = a and y1 cos θ – x1 sin θ = b



as θ is a variable quantity; we eliminate θ. Squaring and adding above equation (we get).

x12 + y1

2 = a2 + b2

⇒ locus of p is x2 + y2 = a2 + b2, which is concentric with given circles.

Method 2:

OAPB is a rectangle

⇒ OP2 = OA2+ AP2

⇒ x12 + y1

2 = a2 + b2

⇒ Locus of P(x1, y1) is x2 + y2 = a2 + b2

Example 9:

The circle x2 + y2 = 1 cuts x-axis at P and Q. Another circle with centre at Q and

variable radius intercepts the first circle at R above x-axis and the line segment PQ at S.

Find the maximum area of the triangle QSR.

Solution:

Method 1.

Equation of circle centred at Q is (x + 1)2 + y2 = µ2

Since point R (cos θ, sin θ) lies on this circle

⇒ µ2 = (cosθ + 1)2 + (sin θ)2 = 2 + 2 cos θ



= 2 (2cos2 θ/2) ⇒ µ = 2 cos θ/2

A = Area of ∆QSR = 1/2 base × altitude

= 1/2 µ sin θ

= cos θ/2 sin θ…… (i)

dA/dθ = cos θ cos θ/2-1/2 sin θ/2 sin θ

Pause:

We will study maxima and minima in module 5.

For max./min. dA/dθ = 0

⇒ tan θ = 2 cot θ/2

⇒ 2t/(1-t2 )=2/t (where t=tan θ/2)

⇒ t = tan θ/2=1/√2

(d2 A)/(d2 θ ) = – sin θ cos θ/2-1/2 cos θ sin θ/2-1/2 sin θ/2 cos θ – 1/4 cos θ/2 sin θ

(A"(θ )(at tan θ/2=1/√2) ) = – ve

From (i)

Maximum area = cos θ/2 sin θ

= 2 sin θ/2 (cos θ/2)2

4/(3√3) sq. units. (Ans.)

Method 2.



Equation of circle I is

x2 + y2 = 1 It cuts x-axis

where P ≡ (1, 0) and

Q ≡ (–1, 0)

Let QR = µ then equation of the circle II.

Centred at Q(–1, 0) and radius = µ is given by

(x + 1)2 + y2 = µ2 …… (1)

Solving it with x-axis; we get S ≡ (µ – 1, 0).

Also solving the two circles, we get the co-ordinates of R

as [(µ2/2)-1,µ/2 √((4-µ2 ) )]

The area of ∆QRS = 1/2 QS × RL

= 1/2 µ µ/2 √((4-µ2 ) )

= A (say)

Now A is max./min. means A2 is max./min.

Let A2 = Z.

Then Z = µ4/16 (4 – µ)2

⇒ dz/dµ=1/4. 4µ3 – (6µ5)/16

(d2 z)/dµ2 = 3µ3 – 30 (µ4/16)

For max./min. of A i.e. max./min. of A2 or Z. we get

dZ/dµ = 0 ⇒ µ = √(8/3) and then



(d2 z)/dµ2 = 3.8/3-30/10.64/9 = –ve

For µ = √((8/3))

Thus area is max when µ = √((8/3))

Also max. area of ∆QRS

1/4.8/3 √((4/3))=(4/3√3) sq. units

Example 10:

Distances from the origin to the centres of three circles x2 + y2 – 2λx = c2(where c is

constant and λ is variable) are in G.P. Prove that the lengths of tangents drawn from any

point on the circle x2 + y2 = c2 to the three circles are in G.P.

Solution:

The equation of the three circles is

x2 + y2 – 2λx = c2 …… (1)

where λ = λ1, λ2, λ3. Their centres are:

(λ 1, 0), (λ2, 0) and (λ3, 0)

Now distances of these points from the origin are λ1, λ2 and λ3

⇒ λ1λ3 = λ22 …… (2)

Now, let P(h, k) be any point on the circle

x2 + y2 = c2, then h2 + k2 = c2 …… (3)

If r1, r2, r3 are the lengths of the tangents from P(h, k) on

The three circles, we then obtain

r12 = h2 + k2 – 2λ1h – c2 = c2 – 2λh – c2, using (3)

r12 = – 2λh



Similarly r22 = – 2λ2h

r32 = – 2λ3h

= r12 r3

2 = 4λ1 λ2 h2 = 4λ22 h2 = (r22)2

= r2r3 – r22 ⇒ r1, r2, r3 are in G.P. (Proved).

Example 11:

Show that the equation x2 + y2 – 4x – ky – 5 = 0 represents (for variable k) a family

of circles passing through two fixed points A and B. Find the equation of the circle belonging

to this family and cutting circle x2 + y2 – 6x – 5y = 0 at right angles.

Solution:

x2 + y2 – 4x – ky – 5 = 0

(x2 + y2 – 4x – 5) + k (– y) = 0

This is the equation of family of circles passing though the intersection point of x2+ y2 –

4x – 5 = 0 (a circle) and a straight line putting y = 0, in x2 + y2 – 4x – 5 = 0 gives

x2 – 4x – 5 = 0

⇒ x = –1, 5

Hence the given circle passes through two fixed points (–1, 0) and (5, 0)

For given family of circles

x2 + y2 – 4x – ky – 5 = 0 …… (1)

g = –2, f = – k/2, c = –5

One member of family (1) and circle x2 + y2 – 6x – 5y – 0 …… (2)

Intersect orthogonally

For circle (2)

g’ = – 3, f’ = –5/2, c’ = 0



For two circle to be orthogonal

2(gg’ + ff’) = (c + c’)

2[(–2) (–3) + (–k/2)(5/2)] = – 5 + 0

2 [6+5K/4] = – 5

12 + 5K/2 = – 5

5K/2 = – 17

K = 34/5

Required equation of circle is

x2 + y2 – 4x + 34/5 y – 5 = 0 (Ans.)

Example 12:

Lines 5x + 12y – 10 = 0 and 5x – 12y – 40 = 0 touch a circle C1 of diameter 6. If the

centre of C1 lies in first quadrant, find the equation of circle C2 which is concentric with

C1 and cuts intercept of length 8 on these lines.

Solution:

Recall:

If a circle touches two lines L1 and L2 then the centre of the circle lies on the angle

bisectors of the lines.

Angle bisector of given lines are

(5x+12y-10)/13=±(5x-12y-40)/13

Taking +ve sign: y = (-5)/4

Taking –ve sign: x = 5

⇒ Given lines L1 and L2 intersect at (5,-5/4)

Since the centre of C1 lies in the first quadrant, it can lie on x = 5 only.



Let the centre of C1 be (5, y1)

⇒ |(5(5)+12y1-10)/√(52+122 )| = 3

⇒ 15 + 12y1 = ± 39

⇒ y1 = 2 or y1 = -54/12 (Neglect)

⇒ Centre of C1 is (5, 2)

Since C2 is concentric with C1, its centre is also (5, 2)

C2 cuts intercept of length 8 on lies

5x + 12y – 10 = 0 and 5x – 12y – 40 = 0

⇒ AB = CD = 8

Recall:

Perpendicular from centre bisects the chord.

⇒ AN = 4

C2 N = 3 (given)

Radius of C2 = r (say)

r2 = (AN)2 + (C2N)2 = 16 + 9

⇒ r = 5



Equation of C2 is (x – 5)2 + (y – 2)2 = 25

Example 13:

Find the equation of the circle which passes through the point (2a, 0) and whose radical

axis with respect to the circle x2 + y2 = a2 is the lines x = a/2.

Solution:

Recall:

Radical axis of two circles is S1 – S2= 0.

Let equation of circle is

x2 + y2 + 2gx + 2fy + c = 0 …… (i)

radical axis of circle (i) and circle x2 + y2 – a2 = 0 is given by

x2 + y2 + 2gx + 2fy + c = x2 + y2 – a2

2gx + 2fy + c + a2 = 0 …… (ii)

It is given that radical axis is x – a/2 = 0 …… (iii)

Comparing (ii) and (iii)

We get f = 0, (c+a2)/2g=-a/2

⇒ ag + a2 + c = 0 …… (iv)

Circle (i) passes through (2a, 0)

4ag + 4a2 + c = 0 …… (v)

From (iv) and (v)

3ag + 3a2 = 0

⇒ g = –a



⇒ c = – (ag + a2) from (iv)

= –(–a2 + a2) = 0

Equation of circle is

x2 + y2 – 2ax = 0 (Ans.)

Example 14:

Show that x2 + y2 + 4y – 1 = 0, x2 + y2 + 6x + y + 8 = 0 and x2 + y2 – 4x – 4y – 37

= 0 touch each other.

Solution:

S1 ≡ x2 + y2 + 4y – 1 = 0

S2 ≡ x2 + y2 + 6x + y + 8 = 0

S3 ≡ x2 + y2 – 4x – 4y – 37 = 0

C1 ≡ (0, –2), C2 ≡ (–3, –1/2), C3 ≡ (2, 2)

r1 = √(4+1) = √5

r2 = √(9+1/4-8)=√5/2

r3 = √(4+4+37)=√45=3√5

C2C3 = √((2+3)2+(2+1/2)2 )=√(25+25/4)=5/2 √5

C1C3 = √((2-0)2+(2+2)2 )=2√5



r1 + r2 = 3/2 √5 = C1C2

r3 – r2 = 5/4 √5 = C2C3

r3 – r1 = 2√5 = C1C3

Recall:

Two circles touch each other.

(i) externally if C1C2 = r1 + r2

(ii) internally if C1C2 = r1 – r2

Example 15:

Find the four common tangents to the circles x2 + y2 – 22x + 4y + 110 = 0 and x2 +

y2 – 22x – 4y – 100 = 0

Solution:

S1 ≡ x2 + y2 – 22x + 4y + 100 = 0

S2 ≡ x2 + y2 + 22x – 4y – 100 = 0

C1 (11, –2), C2 (–11, 2)



r1 = √((11)2+(-2)2-100) = 5

r2 = √((-11)2+(2)2+100) = 15

Out of four common tangents are transverse tangents and other two direct tangents. (1)

and (2) are direct common tangents while (3) and (4) are transverse common tangents.

Recall:

Transverse common tangents divide line joining centres in ratio of radii internally while

direct tangents divides line joining centres in ratio of radii externally.

Let T1, T2 divide C1C2 in ratio of r1 : r2 internally and externally respectively.

Co-ordinates of T1 are (15×11+5×(-11))/(15+5) and (15×(-2)+5×2)/(15+5)

That is T1 is the point (11/2,-1)

Co-ordinates of T2 are (15×11-5(-11))/(15-5) and (15×(-2)-5×2)/(15-5)

that is T2 is the point (22, – 4)

Let the equation to either of the tangents, passing through T1 be

y + 1 = m (x – 11/2) …… (A)

Then the perpendicular from the point (11, –2) on it is equal to + 5 and hence

(m(11-11/2)-(-2+1))/√(1+m2 ) = ± 5

On solving, we have m = -24/7 or 4/3



The required tangents through T1 are therefore

24x + 7y = 125, and 4x – 3y = 25

Similarly the equation to the tangents through T2 is

y + 4 = m (x – 22) …… (B)

where (m(11-22)-(-2+4))/√(1+m2 ) = ± 5

On solving, we have m = 7/24 or –3/4

On substitution in (B) the required equations are therefore

x – 24y = 250 and 3x + 4y = 50

The four common tangents are therefore found. (Ans.)

Example 16:

The circle x2 + y2 – 4x – 4y + 4 = 0 is inscribed in a triangle which has two of its sides

along the co-ordinate axes. The locus of the circumcentre of the triangle is x + y – xy + k

(x2 + y2)1/2 = 0 Find k?

Solution:

Method 1.

The equation of the incircle can be put in the from (x – 2)2 + (y – 2)2 = 4

This implies that the inradius r = 2 …… (i)

Let the hypotenuse of the triangle meet OX and OY at A (a, 0) and B (0, b) respectively.

r = (Area of ∆AOB) / ((1/2)(Sum of sides of ∆AOB)) = ((1/2)ab) /

((1/2)(a+b+√(a2+b2 )))

∴ ⇒ 2 = ab/(a+b+√(a2+b2 ))

Let M (x1, y1) be the circumcentre of ∆OAB. Since ∆ OAB is right angled, it’s circumcentre



is the mid point of hypotenuse.

∴ ⇒ 2 = (4x1 y1) / (2x1+2y1+2√(x12+y1

2 ))

⇒ x1 + y1 + √(x12+y1

2 ) = x1y1

Hence the equation of the locus of M(x1, y1) is

x + y – xy + √(x2+y2 ) = x1y1

Comparing it with the given equation of the locus, we find that

k = 1. (Ans.)

Method 2.

Equation of AB is x/a + y/b = 1 …… (1)

(x1, y1) ≡ (a/2,b/2), where M(x1, y1) is the circumcentre of ∆OAB i.e. midpoint of the

hypotenuse.

(1) becomes: x / (2x1 )+y / (2y1 ) = 1 …… (2)

⇒ In circle (x – 2)2 + (y – 2)2 = 4 touches line (2)

⇒ (2/(2x1 )+2 / (2y1 )-1) / √((1/2x1 )2+ (1/2y1 )

2 ) = –2

⇒ x1 + y1 – x1 y1 + √(x12+y1

2 ) = 0

∴ Locus of M(x1, y1) is

x + y – xy + √(x2+y2 ) = 0

Comparing it with the given equation of the locus, we find that k = 1.

Note:

Distance from (2, 2) to the line x/2x1 +y/2y1 -1 = 0 has been taken –2, because origin

and this point lies on the same side of the origin.

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Solved Examples On Electrochemistry - askIITians...Find the equation of circle having the lines x 2 + 2xy + 3x + 6y = 0 as its normals and having size just sufficient to contain the