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. 

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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 

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

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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.

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

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

​$$\underline{r=5}$$​​

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

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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.

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 $$\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = \underline{5\sqrt{3}}$$​