How do I find the equation of a circle, given radius and centre coordinates? Say I am asked to find, in expanded form without brackets, the equation of a circle with radius 6 and centre 2,3 - how would I go on about doing this?
I know the equation of a circle is $x^2 + y^2 = r^2$, but what do i do with this information?
 A: The equation of a circle with centre $(a,b)$ and radius $r$ is $(x-a)^2+(y-b)^2=r^2$.
A: The equation of a circle with the centre at $(0,0)$ is $x^2+y^2=r^2$. This is because the circle with the radius $r$ is composed of all points which are $r$ away from $(0,0)$, and since the distance of a point $(x,y)$ from $(0,0)$ is $\sqrt{x^2+y^2}$, this means that the equation will be $$\sqrt{x^2+y^2}=r,$$
or, squaring that, $$x^2+y^2=r^2.$$
Hint
To center the point around an arbitrary point, think about how you would calculate the distance between $(x,y)$ and that arbitrary point.
A: The equation gets like this: $$(x-2)^2+(y-3)^2=36,$$which can bee seen as translating a circle with radius 6 and center $(0,0)$ (the equation you mentioned) to the new center $(2,3)$.
A: The general solution for the circle with centre $(a,b)$ and radius $r$ is
$$
(x-a)^2+(y-b)^2=r^2.
$$
Now, we have the centre $(2,3)$ and the radius $6$, therefore the equation of the circle is
\begin{align}
(x-2)^2+(y-3)^2&=6^2\\
x^2-4x+4+y^2-6x+9&=36\\
x^2+y^2-4x-6y+4+9-36&=0\\
\large\color{blue}{x^2+y^2-4x-6y-23}&\large\color{blue}{=0}.
\end{align}
