A circle is defined to be all the points a given distance from the given centre point.

The equation of a circle with radius $r$ and centred at the origin $(0,0)$ is

$\qquad x^2+y^2 = r^2$

Using knowledge of transformations, if we move this circle $a$ units in the $x$ direction and $b$ units in the $y$ direction, so that the circle's centre becomes $(a,b)$, we have the equation:

$\qquad (x-a)^2 + (y-b)^2 = r^2$

When given an equation of the form $x^2 + ax + y^2 + by +c = 0$, it can be written in one of the forms above, by completing the square for both the $x$ part and $y$ part.

Worked Examples

Example 1

Find the equation of the circle with centre $(3, -4)$ and radius $r=6$.

Solution

Using the equation above, $(x-a)^2+(y-b)^2 = r^2$, and the coordinate $(3,-4)$, we have $a=3$ and $b=-4$:

\[(x-3)^2+(y+4)^2=36\]

Example 2

Find the centre and radius of the circle with equation $x^2-4x+y^2-6y+9=0$.

Solution

To get this into the form above where it is easy to read off the centre and radius, complete the square for both the $x$ part and $y$ part.

So the centre is at $(2,3)$ and the radius is $r = \sqrt{4} = 2$.

Example 3

Given the point $P = (3,4)$, work out if this point is inside, outside or on the circle with equation $(x+2)^2+(y-3)^2=9$.

Solution

First, determine the centre and radius of the circle.

Centre $C=(-2,3)$ and radius $r=3$.

If the distance $d$ between point $P$ and the centre $C$ is less than the radius then point $P$ is inside the circle. If the distance equals the radius the point $P$ is on the circle and if the distance is greater than the radius the point $P$ is outside the circle.

The distance $d$ is calculated by Pythagoras' theorem $d^2=(x_2-x_1)^2+(y_2-y_1)^2$