# How do you find the equation of the circle given Center at (0, 8); r = 8?

Jun 14, 2016

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

#### Explanation:

Let the point on te circle be $\left(x , y\right)$.

As its distance from center $\left(0 , 8\right)$ is always $8$, this being a circle with radius $8$, we have

$\sqrt{{\left(x - 0\right)}^{2} + {\left(y - 8\right)}^{2}} = 8$ or

${\left(x - 0\right)}^{2} + {\left(y - 8\right)}^{2} = 64$ or

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

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

graph{x^2+y^2-16x=0 [-12, 28, -10.08, 9.92]}