# What is the equation of the circle with endpoints of the diameter of a circle are (7,4) and (-9,6)?

May 25, 2016

${\left(x + 1\right)}^{2} + {\left(y - 5\right)}^{2} = 65$

#### 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 coords of the centre and r ,the radius.

We require to know the centre and radius to establish the equation.

Given the coords of the endpoints of the diameter , then the centre of the circle will be at the mid-point.

Given 2 points $\left({x}_{1} , {y}_{1}\right) \text{ and } \left({x}_{2} , {y}_{2}\right)$ then the mid-point is.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{1}{2} \left({x}_{1} + {x}_{2}\right) , \frac{1}{2} \left({y}_{1} + {y}_{2}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The mid-point of (7 ,4) and (-9 ,6) is therefore.

$= \left(\frac{1}{2} \left(7 - 9\right) , \frac{1}{2} \left(4 + 6\right)\right) = \left(- 1 , 5\right) = \text{ centre}$

Now the radius is the distance from the centre to either of the 2 endpoints.

Using the $\textcolor{b l u e}{\text{distance formula}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 points}$

The 2 points here are centre (-1 ,5) and endpoint (7 ,4)

$d = \sqrt{{\left(- 1 - 7\right)}^{2} + {\left(5 - 4\right)}^{2}} = \sqrt{65} = \text{ radius}$

We now have centre = (a ,b) = (-1 ,5) and r $= \sqrt{65}$

$\Rightarrow {\left(x + 1\right)}^{2} + {\left(y - 5\right)}^{2} = 65 \text{ is equation of circle}$