Circles: Starter Activity

Show that circle A:

touches circle B:

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CirclesStarter ActivityShow that circle A: touches circle B: First has centre at (-1, 3) with radius Second centre at (2, -3) with radius radius A + radius B = + = 3But distance between centres =

Since distance to centres = sum of radii then

circles must touch ▪

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20

5

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Objectives

The student should be able to :•Find the equation of a tangent to a circle;•Find the equation of a normal to a circle;•Find the equation of a circle through 3 points;and to gain the high grades, to :•Find the length of a tangent from a point;•Find the equation given a chord and tangent;•Prove that a line is a tangent to a circle

Circle from 3 pointsThe best way is to intersect 2 perp bisectors.eg. Find the equation of the circle passing

through A(3, 1), B(8 , 2), and C(2, 6).Grdt AB = so perp grdt = -5

Grdt AC = so perp grdt = 1/5Midpoint AB =

Midpoint AC =

5

1

38

12

12

12

xx

yy

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5

32

16

13

13

xx

yy

),(),(2

3

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),(),(2

7

2

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2

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Circle from 3 points (ctd).So perp bisr of AB is or y = - 5x +

29& Perp bisr of AC is or Hence 2 values of y must be equal

So from which x = 5Subst in the second equation, y = 4So centre at (5, 4) and dist from A to centre isD = So circle is

)x(y2

115

2

3

)x(y2

5

5

1

2

7 3

5

1 xy

35

1295 xx

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Did you spot the short cut ?We have just seen that Grdt AB x Grdt AC = -1 so Angle BAC = 90 degrees !Hence BC is a diameter

Circle from a diametereg. B(8, 2) and C(2 , 6) mark the end points of

the diameter of a circle. Find the equation of the circle.

Diameter BC =

Hence radius =

Centre is midpoint of AB = Equation is then

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),(, 452

62

2

28

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13

Tangents and Normals to circles.

• Find the equation of the tangent and normal to the circle at (10, 11).

First we need the circle in CTS:Well, the grdt of the line joining (10, 11) to the

centre (2, 5) is so grdt of nml = 3/4

Hence, normal is

And tangent is

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4

3

8

6

210

511

)x(y 104

311

)x(y 103

411

Tangents to circles.eg. Find the equation of the tangent to the circle

at the point (1, 5). Well, the grdt of the line joining (1, 5) to the centre is

so grdt of tgt = 2/3Hence, tangent is

eg. For the same circle find the length of the tangent from (10, 11).We will need Pythagoras !Distance from (3, 2) to (10, 11) =

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2

3

13

52

)x(y 13

25

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Chordseg. Given that AB is chord of a circle where A is at (1, 3) and B is at

(4,4). The tgt to the circle at A is the line y = 2x + 1. Find the equation of the circle.Normal at A has grdt -½ so eqtn of normal is Or (1) But grdt AB = and midpoint AB = SoSo perp bisector of AB is (2)Hence, So 7 – x = - 6x + 22So 5x = 15 and then x = 3 and so y = 2 so centre is at (3, 2) Radius is distance from (3,2) to A i.eCircle is

)x(y 12

13

3

1

14

34

),(

2

7

2

5)x(y2

53

2

7

2

7

2

1 xy

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2

7

2

1 xx

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App. of sim. eqtn. & equal rootseg. Show that the line y = 7x + 10 is a tangent to the

circle and find the point of contact.Well at intersection, Expanding the brackets, So So Discrim = Hence equal roots because discrim = 0so line is tangent and point of contact atx = -b/2a = - 70/50 = -7/5, y = 7(-7/5) + 10 = 1/5

contact at (-7/5, 1/5)

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