# How do you write the equation for a circle endpoints of a diameter at (3,8) and (-7,4)?

May 16, 2016

The equation of circle is ${x}^{2} + {y}^{2} + 4 x - 12 y + 11 = 0$

#### Explanation:

The center of the circle is mid point between $\left(3 , 8\right)$ and $\left(- 7 , 4\right)$ i.e. ((3-7)/2,(8+4)/2 or $\left(- 2 , 6\right)$.

The radius is distance between $\left(- 7 , 4\right)$ and $\left(- 2 , 6\right)$ or $\sqrt{{\left(- 2 + 7\right)}^{2} + {\left(6 - 4\right)}^{2}} = \sqrt{25 + 4} = \sqrt{29}$

Hence equation of circle of radius $\sqrt{29}$ and center $\left(- 2 , 6\right)$ is

${\left(x + 2\right)}^{2} + {\left(y - 6\right)}^{2} = 29$ or

${x}^{2} + 4 x + 4 + {y}^{2} - 12 y + 36 - 29 = 0$ or

${x}^{2} + {y}^{2} + 4 x - 12 y + 11 = 0$