circles © christine crisp objectives to know the equation of a circle (cartesian form) to find the...

CirclesCircles

© Christine Crisp

Objectives• To know the equation of a circle (Cartesian form)• To find the intersection of circles with straight lines• To Find the tangent to a circle • To know three circle theorems• To solve circle problems using these theorems

KeywordsChord, Tangent, bisector,

perpendicular, gradient, semi-circle

The equation of a circle

x

y

O

1

Consider a circle, with centre the origin and radius 1 Let P(x, y) be any point on the

circle

P(x, y )


x

y

O

P(x, y )

1

Consider a circle, with centre the origin and radius 1 Let P(x, y) be any point on the

circle

x

y

By Pythagoras’ theorem for triangle OPM, 122 yx

M


x

y

The equation gives a circle because only the coordinates of points on the circle satisfy the equation.

122 yx

e.g. Since the radius is 1, we can see that the point (1, 0) lies on the circle

(1, 0)x1


e.g. Since the radius is 1, we can see that the point (1, 0) lies on the circle

1= the right hand side (r.h.s.)

So, the equation is satisfied by the point (1, 0)

Substituting (1, 0) in the left hand side (l.h.s.) of the equation 122 yx

22 )0()1( l.h.s.

The equation gives a circle because only the coordinates of points on the circle satisfy the equation.

122 yx


x

y

l.h.s. (0. 5, 0. 5)

x

The equation is NOT satisfied by the point (0.5, 0.5).

The point does not lie on the circle

since

)5.0,5.0(122 yx

22 )5.0()5.0(

25.025.0 5.0

r.h.s.

The point does not lie on the circle.


P(x, y )

x

y

O x

y

M

P(x, y )

x

y

O x

y

M

If we have a circle with centre at the origin but with radius r, we can again use Pythagoras’ theorem

r

222 ryx

We get


x

y

Now consider a circle with centre at the point ( a, b ) and radius r.

x ),( ba

r

P(x, y )

x - a

y - b

2)( ax 2r2)( by Using Pythagoras’ theorem as before:


x

y222 ryx

x ),(a

ab

Another way of finding the equation of a circle with centre ( a, b ) is to use a translation from 222 ryx

b

x

• Translate by :222 ryx

b

aReplace x by (x – a) and y by (y – b) 222222 )()( rbyaxryx


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


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

this gives the square of the distance of the point from the centre of the circle

( 4 , -3 )

( 2, 1 ) x

x20


SUMMARY

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

222 )()( rbyax

• 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


Use

222 )()( rbyax

Exercises1. Find the equation of the circle with centre (-1, 2

) and radius 3. Multiply out the brackets to give your answer in the form

2. Determine whether the point (3,-5) lies on, inside or outside the circle with equation 4)3()2( 22 yx

9)2)(2()1)(1( yyxx94412 22 yyxx044222 yxyx

022 cqypxyx

Solution: Substitute x = 3 and y = 5 in l.h.s.22 )35()23(

41 4 so the point lies outside the circle

Solution: 9)2()1( 22 yxa = 1, b = 2, r = 3


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

0124622 yxyx

Finding the centre and radius of a circle

Solution:

First complete the square for x


01242 yy xx 62

2)( x 3 9

)3)(3()3( 2 xxx

xx 62 9

N.B.

so we need to subtract 9 to get

xx 62


Solution:




01242 yy xx 62

9)3( 2 x


Solution:




9)3( 2 x 4)2( 2 y

Next complete the square for y

01262 xx yy 42


Solution:



9)3( 2 x 4)2( 2 y

Copy the constant and complete the equation


Solution:


0124622 yxyx

012



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


SUMMARY

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.

The equation of a circleExercis

es

Solution: Complete the square for x and y: 0416)4(4)2( 22 yx

16)4()2( 22 yx

Find the centre and radius of the circle whose equation is (a) 048422 yxyx

Centre is ( 2, -4 ) and radius is 4

Solution: Complete the square for x and y: 025.0250)50(9)3( 22 yx

9)50()3( 22 yx

(b) 0250622 yxyx

Centre is and radius is 3)50,3(


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.

circles © christine crisp objectives to know the equation of a circle (cartesian form) to find the...

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