Equation of a circle

In a nutshell

The equation of a circle uses Pythagoras' theorem to relate any point $(x,y)$ on the circumference of the circle to the radius of the circle. From the equation of a circle the centre and the radius of the circle can be determined.

The equation of a circle with centre $(0,0)$

When a circle has centre $(0,0)$ and radius $r$ , the equation of the circle is

$\boxed{{x^2} + {y^2} = {r^2}}$

Example 1

The circumference of a circle with centre $(0,0)$ passes through the point $(3,4)$ . What is the radius of the circle?

Draw a sketch using all the information given in the question.

Maths; Circles; KS5 Year 12; Equation of a circle

Using the equation ${{x^2} + {y^2} = {r^2}}$ , solve for $r$ .

$\begin{aligned}x^2 + y^2 &= r^2 \\3^2 + 4^2 &= r^2 \\25 &=r^2 \\\end {aligned}$

$\underline{r=5}$

The equation of a circle with centre $(a,b)$

When a circle has centre $(a,b)$ and radius $r$ , the equation of the circle is

$\boxed{{(x-a)^2} + {(y-b)^2} = {r^2}}$

Example 2

A circle has the equation ${{(x-5)^2} + {(y+6)^2} = {90}}$ . Write down the radius and centre of the circle, then show that the circle passes through $(2,3)$ .

The centre of the circle is $\underline{(5,-6)}$ .

The radius is $\sqrt{90} = \sqrt{9 \times 10} = \sqrt{9} \times \sqrt{10} = \underline{3\sqrt{10}}$

Now substitute the coordinates $(2,3)$ into the equation of the circle to show that the circle passes through that point.

${{(2-5)^2} + {(3+6)^2} = {90}}$

${{(-3)^2} + {(9)^2} = {90}}$

$9 + 81 = {90}$

Therefore, the point does lie on the circle.

Maths; Circles; KS5 Year 12; Equation of a circle

Expanded form of the equation of a circle

You may also see the equation of a circle in its expanded form.

${{x^2} + {y^2} + 2fx + 2gy +c= 0}$

This circle has centre $(-f,-g)$ and radius $\sqrt{{f^2} + {g^2} - c}$ . To find the centre and radius of a circle with its equation in its expanded form, complete the square for the $x$ and $y$ terms.

Example 3

A circle has the equation ${{x^2} +{y^2} + 6x -14y -17 = 0}$ . Find the centre and the radius of the circle.

First the circle needs to be rearrange into the form ${{(x-a)^2} + {(y-b)^2} = {r^2}}$ .

Rearrange the equation to collect $x$ and $y$ terms. 

${{x^2} + 6x + {y^2} -14y = 17}$

Complete the square for $x$  and $y$  terms.

${({x+3)^2} -9 + {(y-7)^2} -49 = 17}$

Collect all number terms to the right-hand side of the equation.

${({x+3)^2} + {(y-7)^2} = 75}$

The centre is $\underline{(-3,7)}$

The radius is $\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = \underline{5\sqrt{3}}$ 