Question

Find an equation of the circle that satisfies the given conditions.

Endpoints of a diameter are$P(-1,3)\mathrm{and}Q(7,-5)$

Short answer

The equation of the circle is ${(x-3)}^2+{(y+1)}^2=32$.

Step-by-step solution

Step 1. Determining general Equation of circle.

An equation of the circle with center $(h,k)$and radius r is

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

This is called the standard form for the equation of the circle.

Step 2. Determining the center and radius of the circle.

Center is the midpoint of the diameter PQ so by the midpoint formula the center is

$\left(\frac{{\displaystyle -1+7}}{{\displaystyle 2}},\frac{{\displaystyle 3-5}}{{\displaystyle 2}}\right)=(3,-1)$

The radiusis the distance from P to the center, so by the distance formula

$$\begin{gathered} r^2={\left(3+1\right)}^2+{\left(-1,-3\right)}^2 \\ =16+16 \\ =32 \end{gathered}$$

Step 3. Determining the equation of a circle.

Here,$h=3,k=-1,r^2=32$

The equation of the circle is

${(x-3)}^2+{(y+1)}^2=32$