Circle Geometry

Circle Geometry

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

Circle Geometry

Circles, like lines, have their own equations on graphs. However, circle equations are a little different, and the geometry a little more in-depth.

A Level AQA Edexcel OCR

Equation of a Circle

The equation of a circle is

(x-\color{orange}a\color{grey})^{2}+(y-\color{blue}b\color{grey})^{2}=\color{green}r\color{grey}^{2}

This circle has a centre of (\textcolor{orange}{a},\textcolor{blue}{b}) and a radius of \textcolor{green}{r}.

A LevelAQAEdexcelOCR

Rearranging a Circle Equation

Some circle equations are not in the form that gives us the centre and radius straight away. However, we can use completing the square to obtain the form we desire.

Example: Find the centre and radius of the circle x^{2}+y^{2}-6x-4y-12=0

(x^{2}-6x)+(y^{2}-4y)-12=0

(x-3)^{2}-9+(y-2)^{2}-4-12=0

(x-3)^{2}+(y-2)^{2}-25=0

(x-3)^{2}+(y-2)^{2}=25

(x-3)^{2}+(y-2)^{2}=5^{2}

Centre is (3,2) and radius is 5.

A LevelAQAEdexcelOCR

Circle Theorems

There are three circle theorems that are important for circle geometry. They are:

A LevelAQAEdexcelOCR
A Level AQA Edexcel OCR

Example 1: Equation of a Circle

Find the equation of this circle, in the form x^{2}+y^{2}+ax+by+c=0.

[2 marks]

Centre is (12,5).

Radius is 16.

Equation is (x-12)^{2}+(y-5)^{2}=16^{2}

x^{2}-24x+144+y^{2}-10y+25=256

x^{2}+y^{2}-24x-10y+169=256

x^{2}+y^{2}-24x-10y-87=0

A LevelAQAEdexcelOCR

Example 2: Tangent of a Circle

Find the equation of the tangent of the circle (x-1)^{2}+(y-1)^{2}=25 at the point (4,5), in y=mx+c form.

[5 marks]

Tangent is perpendicular to radius, so we will find the gradient of the radius to obtain the gradient of the tangent.

Centre: (1,1)

So the radius passes through (1,1) and (4,5).

\begin{aligned}\text{gradient of radius}=&\dfrac{5-1}{4-1}\\[1.2em]=&\dfrac{4}{3}\end{aligned}

Hence, the tangent has gradient -\dfrac{3}{4}

y-y_{1}=m(x-x_{1})

y-5=-\dfrac{3}{4}(x-4)

y-5=-\dfrac{3}{4}x+3

y=-\dfrac{3}{4}x+8

A LevelAQAEdexcelOCR

Example Questions

i) Centre: (1,2)

 

Radius: \sqrt{9}=3

 

ii) Centre: (-3,1)

 

Radius: \sqrt{25}=5

 

iii) Centre: (19,-21)

 

Radius: \sqrt{81}=9

i) (x-3)^{2}+(y-3)^{2}=4

 

ii) (x+1)^{2}+(y-4)^{2}=16

 

iii) (x+10)^{2}+(y+17)^{2}=361

a) x^{2}+y^{2}+12x+8y+3=0

 

(x+6)^{2}-36+(y+4)^{2}-16+3=0

 

(x+6)^{2}+(y+4)^{2}-49=0

 

(x+6)^{2}+(y+4)^{2}=49

 

 

b) x^{2}+y^{2}-6x-8y+21=0

 

(x-3)^{2}-9+(y-4)^{2}-16+21=0

 

(x-3)^{2}+(y-4)^{2}-4=0

 

(x-3)^{2}+(y-4)^{2}=4

a) Centre: (4,5)

 

Radius: \sqrt{25}=5

 

 

b) To find if a point lies on the circle, substitute it into the equation.

 

i) (7,9):

 

(7-4)^{2}+(9-5)^{2}

 

=3^{2}+4^{2}

 

=9+16

 

=25

 

So (7,9) does lie on the circle.

 

ii) (9,5):

 

(9-4)^{2}+(5-5)^{2}

 

=5^{2}+0^{2}

 

=25+0

 

=25

 

So (9,5) does lie on the circle.

 

iii) (2,2):

 

(2-4)^{2}+(2-5)^{2}

 

=(-2)^{2}+(-3)^{2}

 

=4+9

 

=13

 

So (2,2) does not lie on the circle.

 

 

c) Tangent is perpendicular to radius, so gradient of radius will give gradient of tangent.

 

Radius passes through (4,5) and (0,8)

 

\begin{aligned}\text{gradient of radius}&=\dfrac{8-5}{0-4}\\[1.2em]&=\dfrac{3}{-4}\\[1.2em]&=-\dfrac{3}{4}\end{aligned}

 

Hence:

 

\text{gradient of tangent}=\dfrac{4}{3}

 

Tangent passes through (0,8)

 

y-y_{1}=m(x-x_{1})

 

y-8=\dfrac{4}{3}x

 

 

d) Touching x-axis means y=0

 

0-8=\dfrac{4}{3}x

 

\dfrac{4}{3}x=-8

 

x=\dfrac{3}{4}\times(-8)

 

x=-6

 

Touches x-axis at (-6,0)

Additional Resources

MME

Exam Tips Cheat Sheet

A Level
MME

Formula Booklet

A Level

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