# How do you write an equation for a circle with center (6, –8) that passes through (0, 0)?

Sep 18, 2016

${x}^{2} + {y}^{2} - 12 x + 16 y = 0$.

#### Explanation:

Centre of the circle is $C \left(6 , - 8\right)$ and since the circle passes through

the Origin $O \left(0 , 0\right)$,

the radius $r$ is given by, $r \text{=dist.} C P$

$\therefore {r}^{2} = {\left(6 - 0\right)}^{2} + {\left(- 8 - 0\right)}^{2} = 36 + 64 = 100$

$\therefore \text{ the eqn. is, } {\left(x - 6\right)}^{2} + {\left(y + 8\right)}^{2} = 100$, or,

${x}^{2} + {y}^{2} - 12 x + 16 y = 0$.