# How do you write an equation for a circle given center (-sqrt13,42) and passes through the origin?

Oct 29, 2016

${\left(x + \sqrt{13}\right)}^{2} + {\left(y - 42\right)}^{2} = 1777$

#### Explanation:

The general equation of a sphere is,

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

where the centre is at $\left(a , b\right)$ with a radius of r.

So for this what we are missing is the radius but we have the centre and a point $\left(0 , 0\right)$

the distance between the center and the point is,

$\left(0 , 0\right) - \left(- \sqrt{13} , 42\right)$

$\left(\sqrt{13} , - 42\right)$

Using Pythagoras's theorem this means that the radius is,

$r = \sqrt{{\sqrt{13}}^{2} + {42}^{2}}$

$r = \sqrt{1777}$

so back to the general equation,

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

sub in,

${\left(x + \sqrt{13}\right)}^{2} + {\left(y - 42\right)}^{2} = {\left(\sqrt{1777}\right)}^{2}$

${\left(x + \sqrt{13}\right)}^{2} + {\left(y - 42\right)}^{2} = 1777$