18: Circles, Tangents and Chords © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core Modules.

18: Circles, Tangents 18: Circles, Tangents and Chordsand Chords

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 1: AS Core Vol. 1: AS Core ModulesModules

Circle Problems

Module C1AQA Edexc

el

OCR

MEI/OCR

Module C2

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Circle Problems

x

radius

Some properties of circles may be needed in solving problems. This is the 1st one The tangent to a circle is perpendicular to

the radius at its point of contact

tangent

Tangents to Circles

A line which is perpendicular to a tangent to any curve is called a normal.For a circle, the radius is a normal.

Circle Problems

x

Diagrams are very useful when solving problems involving circlese.g.1 Find the equation of the tangent at the

point (5, 7) on a circle with centre (2, 3)

(2, 3)

(5, 7)x

tangent

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

Method: The equation of any straight line is .

cmxy

1mgradient

mgradient

12

1

mm • Find m

using• Substitute for x, y, and m in to find c. cmxy

• Find using 1m12

12

xx

yym

We need m, the gradient of the tangent.

Tangents to Circles

Circle Problems

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

443

43 xy

c )5(743 c

443

4334 xyor

cxy 43

4

3 m

Circle Problems

e.g.2 The centre of a circle is at the point C (-1, 2). The radius is 3. Find the length of the tangents from the point P ( 3, 0).

xC (-1, 2)

Solution:

212

212 )()( yyxxd

P (3,0)x

3

Method: Sketch!

• Find CP and use Pythagoras’ theorem for triangle CPA

A

222 ACPCAP 11920 AP

tangent

tangent• Use 1 tangent and

join the radius.The required length is

AP.

22 )20())1(3( CP 20416 CP

20

11

Circle ProblemsExercise

s

1. Find the equation of the tangent at

the point A(3, -2) on the circle 1322 yx


the point A(7, 6) on the circle 25)2()4( 22 yx

Ans: 1332 xy

Solutions are on the next 2 slides

Ans: 4543 yx

Circle Problems

1. Find the equation of the tangent at the

point

A(3, -2) on the circle

1322 yx

Solution: Centre is (0, 0).

Sketch!

Equation of

tangent is

2

1323 xy o

r

01332 xy

12

121 xx

yym

Gradient of

radius, 3

21

m

Gradient of

tangent,

12

1

mm

2

3 m

cmxy c2

13c )3(223

gradient x

x(0, 0)

(3, -2)

gradient m

1m

Circle Problems


the point A(7, 6) on the circle 25)2()4( 22 yx

Solution: Centre is (4,

2).

445

43 xy 4543 yx o

r

Gradient of

tangent,

4

32 m

12

1

mm

3

41 m

Gradient of

radius,

12

121 xx

yym

cmxy

c445

c )7(643

gradient

(4 , 2)

(7, 6)x

tangent

mgradient

x

1m

Circle Problems

x

Chords of Circles

The perpendicular from the centre to a chord bisects the chord

chord

Another useful property of circle is the following:

Circle Problems

x

chord

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

)3,4(M x• Find the gradient of the

radius

Method: We need m and c in cmxy

• Complete the square to find the centre

• Find the gradient of the chord

• Substitute the coordinates of M into to find c.

cmxy

Circle Problems

)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



0218622 yxyx

Tip to save time: Could you have got the centre without completing the square?

Circle Problems

(b)x

xchord

)6,2(M

Exercise1. A circle has equation

(a) Find the coordinates of the centre, C. (b) Find the equation of the chord with mid-point (2, 6).

01810222 yxyx

Solution:

(a)

01825)5(1)1( 22 yx8)5()1( 22 yx

Centre is ( 1,

5 ) )5,1(C

1m

m

1 m

Equation of chord

is

cxy )6,2( on the

chord

c 26 c 8

8 xyEquation of chord

is

12

121 xx

yym

12

1

mm

112

561

m

Circle Problems

x

Semicircles

The angle in a semicircle is a right angle

diameter

P

Q

A

B

90APB 90AQB

The 3rd property of circles that is useful is:

Circle Problems

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

2m

1m

Circle Problems

B(-2, 4)2m

diameter

C(1, 2)

1m

A(3, 5)

x

Exercise

1. A, B and C are the points (3, 5), ( -2, 4) and (1, 2) respectively. Show that C lies on the circle with diameter AB.

13

2

2

321

mm

12

12

xx

yym

2

3

2

3

31

521

m

3

2

3

2

)2(1

422

m

Solution:

Gradient of BC

Since AC and BC are perpendicular, C lies on the circle diameter AB.

Gradient of AC

Circle Problems

Circle Problems

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

Circle Problems

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.

Circle Problems

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

Circle Problems

)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



0218622 yxyx

Circle Problems

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

18: Circles, Tangents and Chords © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core Modules.

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