# How do you write an equation for a circle with endpoints of a diameter are at (1,1) and (1, -9)?

May 11, 2016

${\left(x - 1\right)}^{2} + {\left(y + 4\right)}^{2} = {5}^{2}$. Expansion gives ${x}^{2} + {y}^{2} - 2 x + 8 y - 8 = 0$

#### Explanation:

The center bisects the diameter. So it is the midpoint

C((1+1)/2, (1-9)/2))=(1, -4).

The radius a is half the diameter #=(1/2)sqrt(0^2+10^2)=(1/2)10=5.

So, the equation of the circle is ${\left(x - 1\right)}^{2} + {\left(y + 4\right)}^{2} = {5}^{2}$

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