College Algebra, Chapter 2, 2.2, Section 2.2, Problem 60

Find an equation of the circle with endpoints of a diameter at $P(-1,3)$ and $Q(7, -5)$.

Recall that the diameter is twice the radius so by getting the distance between the points using distance formula,


$
\begin{equation}
\begin{aligned}

d_{PQ} =& \sqrt{(-5 -3)^2 + (7 - (-1))^2}
\\
\\
d_{PQ} =& \sqrt{(-8)^2 + (8)^2}
\\
\\
d_{PQ} =& \sqrt{64 + 64}
\\
\\
d_{PQ} =& 8 \sqrt{2} \text{ units}

\end{aligned}
\end{equation}
$


Therefore, the radius is..

$\displaystyle r = \frac{d_{PQ}}{2} = \frac{8 \sqrt{2}}{2} = 4 \sqrt{2} $ units

Also, recall that the general equation for the circle with circle $(h,k)$ and
radius $r$ is..


$
\begin{equation}
\begin{aligned}

(x - h)^2 + (y - k)^2 =& r^2
&& \text{Model}
\\
\\
(x - h)^2 + (y - k)^2 =& (4 \sqrt{2})^2
&& \text{Substitute the given}
\\
\\
(x - h)^2 + (y - k)^2 =& 32
&&

\end{aligned}
\end{equation}
$


To get the center $(h,k)$, we get the midpoint of the endpoints of the diameter $PQ$

$\displaystyle h = \frac{-1 + 7}{2} = 3 \text{ and } k = \frac{3 - 5}{2} = -1$

Thus, the equation of the circles..


$
\begin{equation}
\begin{aligned}

(x - 3)^2 + (y - (-1))^2 =& 32
\\
\\
(x - 3)^2 + (y + 1)^2 =& 32


\end{aligned}
\end{equation}
$