# How do you write the equation for a circle with center (-1, -3) and passing through (-2, 0)?

Apr 30, 2016

${\left(x + 1\right)}^{2} + {\left(y + 3\right)}^{2} = 10$

#### Explanation:

Recall that the general formula for a circle is:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where
$x =$x-coordinate
$y =$y-coordinate
$a =$x-coordinate of circle's centre
$b =$y-coordinate of circle's centre
$r =$radius

Using the formula, substitute the circle's centre, $\left(- 1 , - 3\right)$.

${\left(x + 1\right)}^{2} + {\left(y + 3\right)}^{2} = {r}^{2}$

Plug in the point, $\left(- 2 , 0\right)$.

${\left(- 2 + 1\right)}^{2} + {\left(0 + 3\right)}^{2} = {r}^{2}$

Solve for $r$.

$r = \sqrt{{\left(- 2 + 1\right)}^{2} + {\left(0 + 3\right)}^{2}}$

$r = \sqrt{1 + 9}$

$r = \sqrt{10}$

Rewrite the equation.

${\left(x + 1\right)}^{2} + {\left(y + 3\right)}^{2} = {\left(\sqrt{10}\right)}^{2}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\left(x + 1\right)}^{2} + {\left(y + 3\right)}^{2} = 10} \textcolor{w h i t e}{\frac{a}{a}} |}}}$