How do you write an equation of a circle with endpoints of the diameter at (4, -3) and (-2, 5)?

Jun 22, 2016

${\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} = 25$

Explanation:

The standard form of the equation of a circle is.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where (a ,b) are the coordinates of the centre and r, the radius.

We require to find the centre and the radius.

Since we are given the endpoints of the diameter then the centre will be at the mid-point.
The radius will be the distance from the centre to either of the endpoints of the diameter.

To find the centre use the $\textcolor{b l u e}{\text{mid-point formula}}$

Centre $= \left[\frac{1}{2} \left(4 - 2\right) , \frac{1}{2} \left(- 3 + 5\right)\right] = \left(1 , 1\right)$

To find the radius (r) use the $\textcolor{b l u e}{\text{distance formula}}$

Using points (1 ,1) and (4 ,-3) then.

$r = \sqrt{{\left(4 - 1\right)}^{2} + {\left(- 3 - 1\right)}^{2}} = \sqrt{25} = 5$

We now have the centre (1 ,1) and r = 5

Substitute these values into the standard equation.

$\Rightarrow {\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} = 25 \text{ is the equation of the circle }$