Chapter 6 – Circles
The line segment AB is a diameter of a circle, where A and B are (−3, 8) and
(5, 4) respectively.
Find the coordinates of the centre of the circle.
Example 1 – Finding the centre of a circle
The line segment PQ is a diameter of the circle centre (2, −2). Given that P
is (8, −5), find the coordinates of Q.
Example 2 – Problem solving with midpoints
The line segment AB is a diameter of the circle centre C, where A and
B are (−1, 4) and (5, 2) respectively. The line l passes through C and is
perpendicular to AB. Find the equation of l.
Example 3 – Problem solving with lines and circles
Write down the equation of the circle with centre (5, 7) and radius 4.
Example 4 – Finding the equation of a circle
PEUK A2339f | Version 1.0 | Pearson Edexcel | Sep 2020 | DCL1: Public
A circle has equation (x − 3)2 + (y + 4)2 = 20.
a. Write down the centre and radius of the circle.
b. Show that the circle passes through (5, −8).
Example 5 – The equation of a circle
The line segment AB is a diameter of a circle, where A and B are (4, 7)
and (−8, 3) respectively. Find the equation of the circle.
Example 6 – Finding the equation of a circle
Find the centre and the radius of the circle with the equation
x2 + y2 − 14x + 16y − 12 = 0.
Example 7 – Finding the equation of a circle
The circle C has equation (x − 2)2 + (y − 6)2 = 100.
a. Verify that the point P(10, 0) lies on C.
b. Find an equation of the tangent to C at the point (10, 0), giving your
answer in the form ax + by + c = 0.
Example 10 – Finding the equation of a tangent to a circle
A circle C has equation (x − 5)2 + (y + 3)2 = 10.
The line l is a tangent to the circle and has gradient −3.
Find two possible equations for l, giving your answers in the form y = mx + c.
Example 11 – Problem solving using tangents and circles
Find the coordinates of the points where the line y = x + 5 meets the circle
x2 + (y − 2)2 = 29.
Example 8 – Intersection of linear graphs and circles
Show that the line y = x − 7 does not meet the circle (x + 2)2 + y2 = 33.
Example 9 – Problem solving with lines and circles
The points P and Q lie on a circle with centre C, as shown in the diagram.
The point P has coordinates (−7, −1) and the point Q has coordinates (3, −5).
M is the midpoint of the line segment PQ.
The line l passes through the points M
and C.
a. Find an equation for l.
Given that the y-coordinate of C is −8,
b. show that the x-coordinate of C is −4
c. find an equation of the circle.
Example 12 – Problem solving using chords and circles
125
Circles
This is the other possible equation for the tangent.
So the tangents will intersect the circle at
(8, −2) and (2, −4)
y − y1 = m(x − x1)
y + 2 = −3(x − 8)
y = −3x + 22
y − y1 = m(x − x1)
y + 4 = − 3(x − 2)
y = −3x + 2
This is one possible equation for the tangent.
Substitute
(
x 1 , y 1
)
= (2, −4) and m = −3 into the
equation for a straight line.
Substitute (x1, y1) = (8, −2) and m = −3 into the
equation for a straight line.
Example 12
The points P and Q lie on a circle with centre C, as shown in the diagram.
The point P has coordinates (−7, −1) and the point Q has coordinates (3, −5).
M is the midpoint of the line segment PQ.
The line l passes through the points M and C.
a Find an equation for l.
Given that the y-coordinate of C is −8,
b show that the x-coordinate of C is −4
c find an equation of the circle.
y
x
Q(3, –5)
O
P(–7, –1)
l
C
M
a The midpoint M of line segment PQ is:
(
x 1 + x 2
______
2 , y 1 + y 2
______
2
)
=
(
−7 + 3
______
2 , −1 +
(
−5
)
________
2
)
= (−2, − 3)
Gradient of PQ = y 2 − y 1
______
x 2 − x 1 =
−5 − (−1)
________
3 − (−7)
= −4
___
10 = − 2
__
5
The gradient of a line perpendicular to
PQ is 5
__
2
y − y1 = m(x − x1)
y + 3 = 5
__
2 (x + 2)
y + 3 = 5
__
2 x + 5
y = 5
__
2 x + 2
b y = 5
__
2 x + 2
−8 = 5
__
2 x + 2
5
__
2 x = −10
x = −4
Use the midpoint formula with (x1, y1) = (−7, −1)
and (x2, y2) = (3, −5).
Use the gradient formula with (x1, y1) = (−7, −1)
and (x2, y2) = (3, −5).
Substitute (x1, y1) = (−2, −3) and m = 5
_
2 into the
equation of a straight line.
Simplify and leave in the form y = mx + c.
If a gradient is given as a fraction, you can find
the perpendicular gradient quickly by turning the
fraction upside down and changing the sign.
Problem-solving
The perpendicular bisector of any chord passes
through the centre of the circle. Substitute y = −8
into the equation of the straight line to find the
corresponding x-coordinate.
Solve the equation to find x.
The points A (−8, 1), B (4, 5) and
C (−4, 9) lie on the circle, as shown
in the diagram.
a. Show that AB is a diameter of
the circle.
b. Find an equation of the circle.
Example 13 – Problem solving using circle theorems
129
Circles
Example 13
The points A(−8, 1), B(4, 5) and C(−4, 9) lie on the
circle, as shown in the diagram.
a Show that AB is a diameter of the circle.
b Find an equation of the circle.
y
x
C(–4, 9)
O
A(–8, 1)
B(4, 5)
a Test triangle ABC to see if it is a right-
angled triangle.
AB2 = (4−(−8))2 + (5 − 1)2
= 122 + 42 = 160
AC2 = (−4−(−8))2 + (9 − 1)2
= 42 + 82 = 80
BC2 = (−4 − 4)2 + (9 − 5)2
= (−8)2 + 42 = 80
Now, 80 + 80 = 160 so AC2 + BC2 = AB2
So triangle ABC is a right-angled triangle
and AB is the diameter of the circle.
b Find the midpoint M of AB.
(
x 1 + x 2
______
2 , y 1 + y 2
______
2
)
=
(
−8 + 4
______
2 , 1 + 5
_____
2
)
= (−2, 3)
The diameter is
√
____
160 = 4
√
___
10
The radius is 2
√
___
10
(x − a)2 + (y − b)2 = r2
(x + 2)2 + (y − 3)2 =
(
2
√
___
10
)
2
(x + 2)2 + (y − 3)2 = 40
Use d2 = (x2 − x1)2 + (y2 − y1)2 to determine the
length of each side of the triangle ABC.
Use Pythagoras’ theorem to test if triangle ABC
is a right-angled triangle.
If ABC is a right-angled triangle, its longest
side must be a diameter of the circle that passes
through all three points.
For a right-angled triangle, the hypotenuse of the triangle is a
diameter of the circumcircle.
You can state this result in two other ways:
■ If ∠ PRQ = 90° then R lies on the circle with diameter PQ.
■ The angle in a semicircle is always a right angle.
To find the centre of a circle given any three points on the circumference:
■ Find the equations of the perpendicular bisectors
of two different chords.
■ Find the coordinates of the point of intersection of
the perpendicular bisectors.
PQ
R
P
erpendicular
bisect
ors intersect
at the cent
re of
the cir
cle.
The centre of the circle is the midpoint of AB.
Substitute (x1, y1) = (−8, 1) and (x2, y2) = (4, 5).
From part a, A B 2 = 160 .
The radius is half the diameter.
Substitute (a, b) = (−2, 3) and r = 2
√
___
10 into the
equation for a circle.
The points P(3, 16), Q(11, 12) and R(−7, 6) lie on the circumference of a
circle. The equation of the perpendicular bisector of PQ is y = 2x.
a. Find the equation of the perpendicular bisector of PR.
b. Find the centre of the circle.
c. Work out the equation of the circle.
Example 14 – Finding the equation of a circle using chords
