# How do you find the equation of a circle whose diameter has endpoint (8,0) and (0,6)?

Aug 30, 2016

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

#### Explanation:

As the end points of diameter are $\left(8 , 0\right)$ and $\left(0 , 6\right)$, the center is their midpoint i.e. $\left(\frac{8 + 0}{2} , \frac{0 + 6}{2}\right)$ or $\left(4 , 3\right)$.

The distance between $\left(8 , 0\right)$ and $\left(0 , 6\right)$ will be

$\sqrt{{\left(0 - 8\right)}^{2} + {\left(6 - 0\right)}^{2}}$

= $\sqrt{64 + 36}$

= $\sqrt{100} = 10$

Hence diameter is $10$ and radius $\frac{10}{2} = 5$.

As the center is $\left(4 , 3\right)$ and radius is $5$, the equation of circle is

${\left(x - 4\right)}^{2} + {\left(y - 3\right)}^{2} = {5}^{2}$ or

${x}^{2} - 8 x + 16 + {y}^{2} - 6 y + 9 = 25$ or

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