# How do you find the equation of the circle given Center (8,6), r=10?

Jul 14, 2016

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

#### Explanation:

In a circle, a point moves so that it always remains at a fixed distance (called radius) from a point (called center).

Let the point be $\left(x , y\right)$ and as center of circle is $\left(8 , 6\right)$, its distance from this point will always be $10$.

Hence equation of circle is given by

${\left(x - 8\right)}^{2} + {\left(y - 6\right)}^{2} = {10}^{2}$ or

${x}^{2} - 16 x + 64 + {y}^{2} - 12 y + 36 = 100$ or

${x}^{2} + {y}^{2} - 16 x - 12 y + 64 + 36 - 100 = 0$ or

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