# How do you write the equation for a circle whose diameter has endpoint (4,6) and (-2,6)?

Dec 11, 2016

${\left(x - 1\right)}^{2} + {\left(y - 6\right)}^{2} = 9$

#### Explanation:

The equation of a circle is in the form

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

where

Center: $\left(h , k\right)$

radius$= r$

Hence, we need to find the center and the radius given the 2 endpoints of a diameter

A diameter is a chord that passes through the center of the circle.
Since any point on the circle is equidistant from the center, we can say that the center is midway between the two endpoints of a diameter.

We will use the midpoint formula to find the center

$C = M : \left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$

$\implies C : \left(\frac{4 + - 2}{2} , \frac{6 + 6}{2}\right)$

$\implies C : \left(1 , 6\right)$

To find the radius, we will use the distance formula and find the distance between the center and any of the given endpoints of the diameter.

$r = D = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$

$\implies r = \sqrt{{\left(4 - 1\right)}^{2} + {\left(6 - 6\right)}^{2}}$

$\implies r = \sqrt{{3}^{2}}$

#=> r = 3

Hence, the equation of the circle is

${\left(x - 1\right)}^{2} + {\left(y - 6\right)}^{2} = {3}^{2}$

$\implies {\left(x - 1\right)}^{2} + {\left(y - 6\right)}^{2} = 9$