circles © christine crisp objectives to know the equation of a circle (cartesian form) to find the...
TRANSCRIPT
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 )
The equation of a circle
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
The equation of a circle
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
The equation of a circle
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
The equation of a circle
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.
The equation of a 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
The equation of a circle
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:
The equation of a circle
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
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
The equation of a circle
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
The equation of a circle
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
The equation of a circle
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
The equation of a circle
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
The equation of a circle
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
Finding the centre and radius of a circle
Solution:
First complete the square for x
e.g. Find the centre and radius of the circle with equation
The equation of a circle
01242 yy xx 62
9)3( 2 x
Finding the centre and radius of a circle
Solution:
First complete the square for x
e.g. Find the centre and radius of the circle with equation
The equation of a circle
9)3( 2 x 4)2( 2 y
Next complete the square for y
01262 xx yy 42
Finding the centre and radius of a circle
Solution:
e.g. Find the centre and radius of the circle with equation
The equation of a circle
9)3( 2 x 4)2( 2 y
Copy the constant and complete the equation
Finding the centre and radius of a circle
Solution:
e.g. Find the centre and radius of the circle with equation
0124622 yxyx
012
The equation of a circle
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 ,
Finding the centre and radius of a circle
25)2()3( 22 yx
The equation of a circle
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 equation of a circle
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