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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?

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$(x-a)^2 + (y-b)^2 = r^2$ is the general equation for the circle. Hopefully you can recognize what you have to do. – Chinny84 May 29 '14 at 13:15
The search phrase "equation of circle" gives a huge number of hits on Google, many of which immediately answer this question. – user61527 May 30 '14 at 0:01

The equation of a circle with centre $(a,b)$ and radius $r$ is $(x-a)^2+(y-b)^2=r^2$.

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didn't see your answer pop up :). – Chinny84 May 29 '14 at 13:16

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, $$x^2+y^2=r^2.$$


To center the point around an arbitrary point, think about how you would calculate the distance between $(x,y)$ and that arbitrary point.

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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 $(2,3)$.

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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}

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