# 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

## 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 Level

## 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 Level

## Circle Theorems

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

A Level
A Level

## 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 Level

## 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]

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 Level

## 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.

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 $(-6,0)$

A Level

A Level

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