# How do you find the center and radius of the circle given (x-sqrt5)^2+y^2-8y=9?

Dec 6, 2017

Centre : $\left(\sqrt{5} , 4\right)$ and radius: $5$ unit

#### Explanation:

The center-radius form of the circle equation is

(x – h)^2 + (y – k)^2 = r^2 with at the point $\left(h , k\right)$

and the radius being $r$.

${\left(x - \sqrt{5}\right)}^{2} + {y}^{2} - 8 y = 9$ or

${\left(x - \sqrt{5}\right)}^{2} + {y}^{2} - 8 y + 16 = 9 + 16$ or

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

$\therefore h = \sqrt{5} , k = 4 \mathmr{and} r = 5$

Therefore centre is at $\left(\sqrt{5} , 4\right)$ and radius is $5$ unit [Ans]