Circles
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
Keywords
Chord, Tangent, bisector, perpendicular,
gradient, semi-circle
© Christine Crisp
The equation of a circle
Consider a circle, with centre the origin and radius 1
Let P(x, y) be any point on the circle
y
P(x, y )
1
O
x
The equation of a circle
Consider a circle, with centre the origin and radius 1
Let P(x, y) be any point on the circle
By Pythagoras’
theorem for
triangle OPM,
y
P(x, y )
x2  y2  1
1
y
O
x
M
x
The equation of a circle
2
2
The equation x  y  1 gives a circle because only
the coordinates of points on the circle satisfy the
equation.
e.g. Since the radius is 1, we can see that the
point (1, 0) lies on the circle
y
1
x
(1, 0)
x
The equation of a circle
2
2
The equation x  y  1 gives a circle because only
the coordinates of points on the circle satisfy the
equation.
e.g. Since the radius is 1, we can see that the
point (1, 0) lies on the circle
Substituting (1, 0) in the left hand side (l.h.s.) of
the equation x 2  y 2  1
2
2
l.h.s.
 (1)  (0)
1
= the right hand side (r.h.s.)
So, the equation is satisfied by the point (1, 0)
The equation of a circle
The point
2
2
(0.5, 0.5) does not lie on the circle
x  y  1 since
y
l.h.s.  (0. 5) 2  (0. 5) 2
 0. 25  0. 25
 0. 5
 r.h.s.
(0. 5, 0. 5)
x
x
The equation is NOT satisfied by the point (0.5, 0.5).
The point does not lie on the circle.
The equation of a circle
If we have a circle with centre at the origin but
with radius r, we can again use Pythagoras’ theorem
y
We get
x y r
2
2
P(x, y )
2
r
O
y
x
M
x
The equation of a circle
Now consider a circle with centre at the point ( a, b )
and radius r.
y
P(x, y )
r
y-b
(a , b)
x
x-a
x
Using Pythagoras’ theorem as before:
( x  a ) 2  ( y  b) 2  r 2
The equation of a circle
Another way of finding the equation of a circle with
centre ( a, b ) is to use a translation from x 2  y 2  r 2
y
x2  y2  r 2
(a, b )
x
x
a
b
x
 a 2
2
2
2 2
2
2
2
Replace
and
y
by
(y
–
b)
 y x byr (xx– a)

(
x

a
)

(
y

b
)
• x Translate
by
:
y r
 
b
 r2
The equation of a circle
SUMMARY
The equation of a circle with centre ( a, b ) and
radius r is
( x  a ) 2  ( y  b) 2  r 2
We usually leave the equation in this form
without multiplying out the brackets
The equation of a 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?
Solution: Using the formula, ( x  a ) 2  ( y  b) 2  r 2
the circle is ( x  4) 2  ( y  ( 3)) 2  5 2

( x  4) 2  ( y  3) 2  25
Substituting the coordinates ( 2, 1 ):
l.h.s.  ( 2  4) 2  (1  3) 2
 4  16
 20  25
x
( 2, 1 )
20 x( 4 , -3 )
this gives the square of the
distance of the point from the
centre of the circle
Since the distance of the point from the centre is less
than the radius, the point ( 2, 1 ) is inside the circle
The equation of a circle
SUMMARY
•
The equation of a circle with centre ( a, b )
and radius r is
( x  a ) 2  ( y  b) 2  r 2
•
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 r 2
Exercises
The equation of a circle
1. Find the equation of the circle with centre (-1, 2 ) and
radius 3. Multiply out the brackets to give your
answer in the form x 2  y 2  px  qy  c  0
Solution: Use ( x  a ) 2  ( y  b) 2  r 2
a = 1, b = 2, r = 3 
( x  1) 2  ( y  2) 2  9
 ( x  1)(2 x  1)  ( y 22)( y  2)  9
x  2x  1  y  4 y  4  9


x 2  y 2  2x  4 y  4  0
2. Determine whether the point (3,-5) lies on, inside or
outside the circle with equation ( x  2) 2  ( y  3) 2  4
Solution: Substitute x = 3 and y = 5 in l.h.s.
(3  2) 2  (5  3) 2
 1  4  4 so the point lies outside the circle
The equation of a circle
Finding the centre and radius of a circle
e.g. Find the centre and radius of the circle with
equation
x 2  y 2  6 x  4 y  12  0
Solution:
First complete the square for x
The equation of a circle
Finding the centre and radius of a circle
e.g. Find the centre and radius of the circle with
equation
x 2  y 2  6 x  4 y  12  0
Solution:
( x  3 )2  9
First complete the square for x
N.B.
( x  3) 2  ( x  3)( x  3)
 x 2  6x  9
so we need to subtract 9 to get x 2  6 x
The equation of a circle
Finding the centre and radius of a circle
e.g. Find the centre and radius of the circle with
equation
x 2  y 2  6 x  4 y  12  0
Solution:
( x  3 )2  9
First complete the square for x
The equation of a circle
Finding the centre and radius of a circle
e.g. Find the centre and radius of the circle with
equation
x 2  y 2  6 x  4 y  12  0
Solution:
( x  3 )2  9  ( y  2 )2  4
Next complete the square for y
The equation of a circle
Finding the centre and radius of a circle
e.g. Find the centre and radius of the circle with
equation
x 2  y 2  6 x  4 y  12  0
Solution:
2

(
y

2
)
 4 12  0
(x  3 )  9
2
Copy the constant and complete the equation
The equation of a circle
Finding the centre and radius of a circle
e.g. Find the centre and radius of the circle with
equation
x 2  y 2  6 x  4 y  12  0
Solution:
( x  3 ) 2  9  ( y  2) 2  4  12  0
Finally collect the constant terms onto the r.h.s.
( x  3) 2  ( y  2) 2  25
By comparing with the equation ( x  a ) 2  ( y  b) 2  r 2,
we can see the centre is ( 3, 2 ) and the radius is 5.
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 ( x  a ) 2  ( y  b) 2  r 2
The centre is (a, b) and the radius is r.
The equation of a circle
Exercises Find the centre and radius of the circle whose
equation is (a) x 2  y 2  4 x  8 y  4  0
(b) x 2  y 2  6 x  y  0  25  0
Solution: Complete the square for x and y:
 ( x  2) 2  4  ( y  4) 2  16  4  0
( x  2) 2  ( y  4) 2  16

 Centre is ( 2, -4 ) and radius is 4
Solution: Complete the square for x and y:



( x  3) 2  9  ( y  0  5) 2  0  25  0.25  0
( x  3) 2  ( y  0  5) 2  9
Centre is (3, 0  5) and radius is 3
The equation of a circle
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