1. AS-Level Maths: Core 2 for Edexcel
C2.2 Coordinate geometry
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2. The equation of a circle
Using circle properties
Tangents and normals
Examination-style questions
Contents
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3. The equation of a circle centred on the origin
Suppose we have a circle with its centre on the origin O and radius r.
If P(x, y) is any point on the circle we can write OP = r. x O r y
P(x, y)
Let’s add another point Q on the x-axis so OQ = x and PQ = y. y x Q
Using Pythagoras’ theorem:
OQ2 + PQ2 = OP2
So x2 + y2 = r2
The equation of a circle of radius r centred on the origin is
x2 + y2 = r2

4. Translating circles
Use this activity to demonstrate translations of the circle x2 + y2 = r2.

5. The equation of a circle
Suppose the circle is not centred on the origin but centred on the point C(a, b).
Let P(x, y) be a point on the circle so that CP = r. x y
P(x, y)
The vertical distance from C to P is y – b. r
y – b
C(a, b)
The horizontal distance from C to P is x – a.
x – a
Compare this to translating the circle with equation x2 + y2 = r2 a units to the right and b units up.
Using Pythagoras theorem:
CP2 = (x – a)2 + (y – b)2
So r2 = (x – a)2 + (y – b)2

6. The equation of a circle
The equation of a circle of radius r centred on the point (a, b) is (x – a)2 + (y – b)2 = r2
This equation can be expanded as follows:
(x – a)2 + (y – b)2 = r2
x2 – 2ax + a2
+ y2 – 2by + b2
= r2
Rearranging:
Students should be able to recognize the equation of a circle given in either of the two forms shown.
The second form is often written as x2 + y2 – 2gx – 2fy + c = 0.
x2 + y2 – 2ax – 2by + a2 + b2 – r2 = 0
Since a, b and r are constants, the equation of a circle can be written in the form:
x2 + y2 – 2ax – 2by + c = 0

7. Finding the centre and the radius
Given the equation of a circle, we can find the coordinates of its centre and the length of its radius. For example:
Find the centre and the radius of a circle with the equation (x – 2)2 + (y + 7)2 = 64
By comparing this to the general form of the equation of a circle of radius r centred on the point (a, b):
(x – a)2 + (y – b)2 = r2
Students should also be able to write down the equation of a circle given the coordinates of its centre and the length of its radius.
We can deduce that for the circle with equation
(x – 2)2 + (y + 7)2 = 64
The centre is at the point (2, –7) and the radius is 8.

8. Finding the centre and the radius
When the equation of a circle is given in the form
x2 + y2 – 2ax – 2by + c = 0
we can use the method of completing the square to write it in the form
(x – a)2 + (y – b)2 = r2
For example:
Find the centre and the radius of a circle with the equation x2 + y2 + 4x – 6y + 9 = 0
Start by rearranging the equation so that the x terms and the y terms are together:
x2 + 4x + y2 – 6y + 9 = 0

9. Finding the centre and the radius
We can complete the square for the x terms and then for the y terms as follows:
x2 + 4x =
(x + 2)2 – 4
y2 – 6y =
(y – 3)2 – 9
The equation of the circle can now be written as:
(x + 2)2 – 4 + (y – 3)2 – = 0
(x + 2)2 + (y – 3)2 = 4
(x + 2)2 + (y – 3)2 = 22
The centre is at the point (–2, 3) and the radius is 2.

10. Worked example
Write down the equation of the circle with centre (3, 1) and radius .
Show that the point (4, –1) lies on the circle.
a) The centre is (3, 1) so a = 3 and b = 1 in the equation
(x – a)2 + (y – b)2 = r2
r = and so the equation of the circle is
(x – 3)2 + (y – 1)2 = 5
b) The point (4, –1) lies on the circle if x = 4 and y = –1
satisfies the equation of the circle.
Substituting:
(4 – 3)2 + (–1 – 1)2 = 12 + (–2 )2
= 1 + 4
= 5
 The point (4, –1) lies on the circle.

11. Using circle properties
The equation of a circle
Using circle properties
Tangents and normals
Examination-style questions
Contents
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12. The angle in a semicircle
The angle in a semicircle is a right angle.
Use this activity to demonstrate that the angle subtended by the diameter of a circle is always a right angle. A formal proof of this theorem is not required at this stage.

13. The angle in a semicircle
The three points A(1, 1), B(2, 8) and C(–5, 9) lie on the circle with equation (x + 2)2 + (y – 5)2 = 25.
Show that AC is a diameter of the circle.
Using the angle in a semicircle theorem, if AC is a diameter of the circle then the angle between AB and BC must be a right angle. x y
A(1, 1)
B(2, 8)
C(–5, 9)
In other words, AB and BC must be perpendicular.
mAB =
The converse of the theorem shown on the previous slide is that if the angle subtended by a chord is a right angle then that chord must be the diameter of the circle.
We could also show that AC is a diameter by finding the equation of AC and showing that it passes through the centre of the circle (–2, 5). 7
mBC =
mAB mBC = –1 and so AB and BC are perpendicular and AC is therefore a diameter of the circle.

14. The perpendicular from the centre to a chord
The perpendicular from the centre of a circle to a chord bisects the chord.
The converse of this theorem is also true. That is, that the perpendicular bisector of a cord always passes through the centre of a circle.

15. The perpendicular from the centre to a chord
The points A(–1, 5) and B(4, 4) lie on the circumference of a circle with equation (x – 1)2 + (y – 2)2 = 13 .
Give the equation of a diameter of the circle.
The perpendicular bisector of the chord AB passes through the centre of the circle and is therefore a diameter.
The gradient of the chord AB =
So the gradient of the perpendicular bisector = 5
Before talking through the method for finding the equation of a diameter of the circle, discuss how we can verify that the points (–1, 5) and (4, 4) lie on the circle by substituting them into the equation of the circle.
Also, the perpendicular bisector passes through the mid-point of AB. That is the point given by,

16. The perpendicular from the centre to a chord
We can use y – y1 = m(x – x1) to write the equation of the line that passes through with gradient 5 as
This line is a diameter of the circle (x – 1)2 + (y – 2)2 = 13 if it passes through its centre.
The centre of the circle is the point (1, 2).
Explain that we can check whether y = 5x – 3 is a diameter of the circle by checking whether it passes through the centre of the circle. The equation of the circle is given in the form (x – a)2 + (x – b)2 = r2 where (a, b) is the centre of the circle and so we can find the coordinates of the centre directly from the equation.
So if x = 1 and y = 2 satisfy the equation y = 5x – 3 then it is a diameter of the circle.
RHS = 5(1) – 3
= 2 = LHS as required.

17. Contents Tangents and normals The equation of a circle
Using circle properties
Tangents and normals
Examination-style questions
Contents
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18. Tangents and normals
The normal to a curve at a given point P is perpendicular to the tangent at that point.
In the case of a circle, the normal at any point passes through the centre of the circle.
normal
tangent P
If we know the equation of a circle, we can find the equation of both the tangent and the normal at any given point on the circumference.

19. Find the equation of the tangent and the normal at the point P(6, 2).
Tangents and normals
A circle is given by the equation x2 + y2 – 8x + 4y = 0.
Find the equation of the tangent and the normal at the point P(6, 2).
Rearrange the equation into the form (x – a)2 + (y – b)2 = r2:
x2 + y2 – 8x + 4y = 0
(x – 4)2 + (y + 2)2 – 16 – 4 = 0
(x – 4)2 + (y + 2)2 = 20
We can see from this form of the equation that the centre of the circle is (4, –2).
P(6, 2)
(4, –2)
(4, –2) and (6, 2) lie on the normal.

20. Tangents and normals The gradient of the normal is 2
Using y – y1 = m(x – x1) with this gradient and the point (6, 2) we can write the equation of the normal as:
y – 2 = 2(x – 6)
y – 2 = 2x – 12
y = 2x – 10
Now, the tangent is perpendicular to the normal so its gradient is – . It also passes through (6, 2) so its equation is:
2y – 4 = –x + 6
x + 2y – 10 = 0

21. Examination-style questions
The equation of a circle
Using circle properties
Tangents and normals
Examination-style questions
Contents
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22. Examination-style question
a) Find the equation of the circle which passes through the points (1, 1), (0, 4) and (8, 0).
b) Find the centre and the radius of the circle.
a) Let the equation of the circle be x2 + y2 – 2ax – 2by + c = 0
At (1, 1)
2 – 2a – 2b + c = 0 1
At (0, 4)
16 – 8b + c = 0
c = 8b – 16 2
At (8, 0)
64 – 16a + c = 0
c = 16a – 64 3

23. Examination-style question
Substituting into : 2 1
2 – 2a – 2b + 8b – 16 = 0
– 2a + 6b = 16 4
Substituting into : 2 3
8b – 16 = 16a – 64
b – 2 = 2a – 8
2a – b = 6 5
+ : 4 5
5b = 20
b = 4
a = 5
c = 16
So the equation of the circle is
x2 + y2 – 10x – 8y + 16 = 0

24. Examination-style question
b) The centre and the radius can be found by rearranging the equation of the circle to the form (x – a)2 + (y – b)2 = r2
x2 – 10x + y2 – 8y + 16 = 0
(x – 5)2 – 25 + (y – 4)2 – = 0
(x – 5)2 + (y – 4)2 = 25
 The centre of the circle is (5, 4) and the radius is √25 = 5