# Circle Geometry

## Circle Geometry Revision

**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**.

**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}.

**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.

**Circle Theorems**

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

**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

**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).

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

## Circle Geometry Example Questions

**Question 1: **What are the centre and radius of these circles?

i) (x-1)^{2}+(y-2)^{2}=9

ii) (x+3)^{2}+(y-1)^{2}=25

iii) (x-19)^{2}+(y+21)^{2}=81

**[3 marks]**

i) Centre: (1,2)

Radius: \sqrt{9}=3

ii) Centre: (-3,1)

Radius: \sqrt{25}=5

iii) Centre: (19,-21)

Radius: \sqrt{81}=9

**Question 2: **Write the equations of the circles with the following centres and radii:

i) Centre: (3,3)

Radius: 2

ii) Centre: (-1,4)

Radius: 4

iii) Centre: (-10,-17)

Radius: 19

**[3 marks]**

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

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

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

**Question 3: **Rearrange these equations into the form (x-a)^{2}+(y-b)^{2}=r^{2}:

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

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

**[4 marks]**

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

**Question 4: **Consider the circle (x-4)^{2}+(y-5)^{2}=25.

a) State the centre and radius.

b) Which of these points does the circle pass through?

i) (7,9)

ii) (9,5)

iii) (2,2)

c) Find the equation of the tangent at the point (0, 8).

d) Find the co-ordinates of the point where this tangent touches the x-axis.

**[10 marks]**

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)

## You May Also Like...

### MME Learning Portal

Online exams, practice questions and revision videos for every GCSE level 9-1 topic! No fees, no trial period, just totally free access to the UK’s best GCSE maths revision platform.