# How do you find the equation of a circle passes through the origin and has its center at (-1,-9)?

Jun 20, 2016

${\left(x + 1\right)}^{2} + {\left(y + 9\right)}^{2} = 82$

#### Explanation:

If a circle has a center at $\left({x}_{c} , {y}_{c}\right) = \left(- 1 , - 9\right)$ and passes through the origin (i.e. $\left(0 , 0\right)$ )
then it has a radius of r=$\sqrt{{\left(- 1 - 0\right)}^{2} + {\left(- 9 - 0\right)}^{2}} = \sqrt{82}$

the general formula for a circle with center $\left({x}_{c} , {y}_{c}\right)$ and radius $r$ is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - {x}_{c}\right)}^{2} + {\left(y - {y}_{c}\right)}^{2} = {r}^{2}$

So in this particular case:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - \left(- 1\right)\right)}^{2} + {\left(y - \left(- 9\right)\right)}^{2} = 82$
or
$\textcolor{w h i t e}{\text{XXX}} {\left(x + 1\right)}^{2} + {\left(y + 9\right)}^{2} = 82$

graph{(x+1)^2+(y+9)^2=82 [-24.15, 21.45, -21.76, 1.06]}