# The general form of the equation of a circle is x2+y2−4x−8y−5=0. What are the coordinates of the center of the circle?

May 11, 2017

The center of the circle is at $\left(2 , 4\right)$ and the circle has a radius of $5$.

#### Explanation:

Formatted question: ${x}^{2} + {y}^{2} - 4 x - 8 y - 5 = 0$

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

To find the standard form of the circle, we need to complete the square and move the constants to the right side:
${x}^{2} + {y}^{2} - 4 x - 8 y = 5$
${\left(x - 2\right)}^{2} - 4 + {\left(y - 4\right)}^{2} - 16 = 5$
${\left(x - 2\right)}^{2} + {\left(y - 4\right)}^{2} = 25$

Therefore, the center of the circle is at $\left(2 , 4\right)$ and the circle has a radius of $5$.