# How do you find the equation of the circle which has it centre at point (2,3) and passes through the origin?

Nov 13, 2015

${\left(x - 2\right)}^{2} + {\left(y - 3\right)}^{2} = 13$

#### Explanation:

If the center is at $\left(2 , 3\right)$ and the circle passes through $\left(0 , 0\right)$
then the radius is $r = \sqrt{{2}^{2} + {3}^{2}} = \sqrt{13}$

The general equation of a circle is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - {c}_{x}\right)}^{2} + {\left(y - {c}_{y}\right)}^{2} = {r}^{2}$
where $\left({c}_{x} , {c}_{y}\right)$ is the coordinate center of the circle.

graph{(x-2)^2+(y-3)^2=13 [-6.875, 13.125, -2.88, 7.12]}