# How do you find the center and radius of the circle x^2+y^2-4x+8y-5=0 ?

Aug 25, 2016

Radius: $5$ units
Center: $\left(2 , - 4\right)$

#### Explanation:

Do a double completion of square; the x's with the x's and the y's with the y's.

$1 \left({x}^{2} - 4 x + m - m\right) + 1 \left({y}^{2} + 8 y + n - n\right) - 5 = 0$

$m = {\left(\frac{b}{2}\right)}^{2} \text{ AND } n = {\left(\frac{b}{2}\right)}^{2}$

$m = {\left(- \frac{4}{2}\right)}^{2} \text{ AND } n = {\left(\frac{8}{2}\right)}^{2}$

$m = 4 \text{ AND } n = 16$

$1 \left({x}^{2} - 4 x + 4 - 4\right) + 1 \left({y}^{2} + 8 y + 16 - 16\right) = 0 + 5$

$1 \left({x}^{2} - 4 x + 4\right) - 4 + 1 \left({y}^{2} + 8 y + 16\right) - 16 = 5$

${\left(x - 2\right)}^{2} - 4 + {\left(y + 4\right)}^{2} - 16 = 5$

${\left(x - 2\right)}^{2} + {\left(y + 4\right)}^{2} = 5 + 16 + 4$

${\left(x - 2\right)}^{2} + {\left(y + 4\right)}^{2} = 25$

In the form ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$, the radius is given by $r$ and the centre $\left(a , b\right)$.

Hence, the centre is at $\left(2 , - 4\right)$ and the radius measures $5$ units. The graph of this circle confirms.

Hopefully this helps!