# How do you write an equation for a circle with endpoints of a diameter at the points (7,4) and (-9,6)?

Oct 25, 2016

(x+1)^2+(y-5)^"=65

#### Explanation:

The standard equation of a circle is: (x-a)^2+(y-b)^"=r^2
where $\left(a , b\right)$ are the co-ordinates of the centre and $r$ is the radius.

Endpoints of diameter $\left(7 , 4\right)$ & $\left(- 9 , 6\right)$

The centre is the MID-POINT of a diameter

$\left(a , b\right) = \left(\frac{7 + \left(- 9\right)}{2} , \frac{4 + 6}{2}\right)$

giving $\left(a , b\right) = \left(- 1 , 5\right)$

To find the radius use Pythagoras on one end point of diameter and the centre.

$r = \sqrt{{\left(- 1 - 7\right)}^{2} + {\left(5 - 4\right)}^{2}}$

$r = \sqrt{65}$

so circle equation is;

(x+1)^2+(y-5)^"=65