# How to find the center and radius of x^2 +(y+2)^2=72?

Aug 5, 2018

Center @ $\left(0 , - 2\right)$, radius $6 \sqrt{2}$

#### Explanation:

Recall the equation of a circle

$\overline{\underline{|} \textcolor{w h i t e}{\frac{2}{2}} {\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \textcolor{w h i t e}{\frac{2}{2}} |}$, with center $\left(h , k\right)$ and radius $r$.

With this in mind, we can rewrite our equation as

${\left(x - 0\right)}^{2} + {\left(y - - 2\right)}^{2} = \sqrt{72}$

This tells us that our circle is centered at $\left(0 , - 2\right)$, and our radius is $\sqrt{72}$, which can be simplified as

$\sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36} \sqrt{2} = 6 \sqrt{2}$

We are centered at $\left(0 , - 2\right)$ and our radius is $6 \sqrt{2}$.

Hope this helps!