# What is the center of the circle given by the equation (x + 5)^2 + (y - 8)^2 = 1?

Jan 19, 2016

The centre of the circle is $\left(- 5 , 8\right)$

#### Explanation:

The basic equation of a circle centred on the point $\left(0 , 0\right)$ is ${x}^{2} + {y}^{2} = {r}^{2}$ when $r$ is the radius of the circle.

If the circle is moved away to some point $\left(h , k\right)$ the equation becomes ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

In the given example $h = - 5$ and $k = 8$

The centre of the circle is therefore $\left(- 5 , 8\right)$