Pupils notes for Circle Lessons

The equation of a circle with centre ( a, b ) and radius r is

222 )()( rbyax

We usually leave the equation in this form without multiplying out the brackets

SUMMARY

To determine whether a point lies on, inside, or outside a circle, substitute the coordinates of the point into the l.h.s. of the equation of the circle and compare the answer with 2r

Since the distance of the point from the centre is less than the radius, the point ( 2, 1 ) is inside the circle

e.g. Find the equation of the circle with centre ( 4, -3 ) and radius 5. Does the point ( 2, 1 ) lie on, inside, or outside the circle?

25)3()4( 22 yx

Substituting the coordinates ( 2, 1 ):

l.h.s. 22 )31()42( 164

20 25

Solution: Using the formula, 222 )()( rbyax 222 5))3(()4( yxthe circle is

To find the centre and radius of a circle given in a form without brackets:• Complete the square for the x-terms

• Complete the square for the y-terms

• Collect the constants on the r.h.s.

• Compare with 222 )()( rbyax

The centre is (a, b) and the radius is r.

SUMMARY

e.g. Find the centre and radius of the circle with equation

0124)2(9)3( 22 yx

0124622 yxyx

Finally collect the constant terms onto the r.h.s.

Solution:

we can see the centre is ( 3, 2 ) and the radius is 5.

222 )()( rbyax By comparing with the equation ,

25)2()3( 22 yx

Tangent: 042 acb

No points of intersection:

042 acb

2 points of intersection:

042 acb

SUMMARY

The discriminant of the quadratic equation formed by eliminating y from the equations of a straight line and a circle tells us how the line and circle are related.

The tangent to a circle is perpendicular to the radius at its point of contact

The perpendicular from the centre to a chord bisects the chord

The angle in a semicircle is a right angle

Properties of Circles

Diagrams are nearly always needed when solving problems involving circles.

A line perpendicular to a tangent to any curve is called a normal. The radius of a circle is therefore a normal.

e.g.1 Find the equation of the tangent at the point (5, 7) on a circle with centre (2, 3)

cmxy

Solution:12

121 xx

yym

12

1

mmm

Substitute the point that is on the tangent, (5, 7):

x(2, 3)

(5, 7)x

tangent

1mgradient

mgradient

3

4

25

371

m

4

32 m

443

43 xy

c )5(743 c

443

4334 xyor

cxy 43

)3,4(M

x

chord

x

02116)4(9)3( 22 yx

)4,3(Centre C is

4)4()3( 22 yx)4,3(C

143

341

m12

121 xx

yym

1 m1

21

mm

c 1c )4(131 xy is chord

Solution: 0218622 yxyx

1mm

cmxy

The point M (4, 3) is the mid-point of a chord. Find the equation of this chord.

e.g. A circle has equation 0218622 yxyx

x

e.g. A circle has diameter AB where A is ( -1, 1) and B is (3, 3). Show that the point P (0, 0) lies on the circle.

diameter

A(-1, 1)

B(3, 3)

Method: If P lies on the circle the lines AP and BP will be perpendicular.

Solution:12

12

xx

yym

101

011

m

P(0, 0)

Hence and P is on the circle. 90APB

Gradient of AP:

Gradient of BP: 103

032

m

So, . 121 mm

1m

2m