# How do you identify the center and radius of the circle (x+3)^2+(y-8)^2=16?

May 29, 2017

center:(-3, 8); " radius "= 4

#### Explanation:

Given: ${\left(x + 3\right)}^{2} + {\left(y - 8\right)}^{2} = 16$

The standard form of a circle ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$, where center is $\left(h , k\right)$ and $r$ = radius.

From the given equation ${\left(x - - 3\right)}^{2} + {\left(y - 8\right)}^{2} = 16$ to get a positive value, there must be a negative.

This means the center is $\left(- 3 , 8\right)$

Radius $= \sqrt{16} = 4$