# What is the center and radius of the circle with equation x^2 + y^2 – 4x + 22y + 61 = 0?

Jan 3, 2016

The center is at $\left(2 , - 11\right)$ and the radius is $8$.

#### Explanation:

All the manipulations that will be done are in effort to make the equation resemble the general form of a circle:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

Where the center is $\left(h , k\right)$ and the radius is $r$.

Sort the $x$ and $y$ terms and move the constant to the other side.

${x}^{2} - 4 x + {y}^{2} + 22 y = - 61$

Complete the square for the $x$ and $y$ parts. Balance the equation by adding the constants to the left and right sides.

$\left({x}^{2} - 4 x + 4\right) + \left({y}^{2} + 22 y + 121\right) = - 61 + 4 + 121$

${\left(x - 2\right)}^{2} + {\left(y + 11\right)}^{2} = 64$

Thus, the center is at $\left(2 , - 11\right)$ and the radius is $\sqrt{64} = 8$.