# How do you write the equation of the circle with Center (3, 4) and radius 6?

May 7, 2016

${x}^{2} + {y}^{2} - 6 x - 8 y - 11 = 0$

#### Explanation:

We have to find the equation of a circle whose center is $\left(3 , 4\right)$ and radius $6$.

As the distance of any point on the circle from center will always be $6$

${\left(x - 3\right)}^{2} + {\left(y - 4\right)}^{2} = {6}^{2}$ or

${x}^{2} - 6 x + 9 + {y}^{2} - 8 y + 16 = 36$ or

${x}^{2} + {y}^{2} - 6 x - 8 y + 9 + 16 - 36 = 0$ or

${x}^{2} + {y}^{2} - 6 x - 8 y - 11 = 0$