College Algebra, Chapter 2, Review Exercises, Section Review Exercises, Problem 10

Find an equation of the circle that contains the points $P(2,3)$ and $Q(-1,8)$ and has the midpoint of the segment $PQ$ as its center
Recall that the general equation for the circle with center $(h,k)$ and radius $r$ is...
$(x - h)^2 + (y - k)^2 = r^2$

To get the center, we use midpoint formula...

$
\begin{equation}
\begin{aligned}
m_{PQ}, \quad x &= \frac{2-1}{2} = \frac{1}{2}\\
\\
y &= \frac{3+8}{2} = \frac{11}{2}
\end{aligned}
\end{equation}
$

Therefore, $\displaystyle \left( \frac{1}{2}, \frac{11}{2} \right)$
Thus, $\displaystyle \left( x - \frac{1}{2} \right)^2 + \left( y - \frac{11}{2} \right)^2 = r^2$
Since the circle pass through $P(2,3)$

$
\begin{equation}
\begin{aligned}
\left( 2 - \frac{1}{2} \right)^2 + \left( 3 - \frac{11}{2} \right)^2 &= r^2\\
\\
\left( \frac{3}{2} \right)^2 + \left( \frac{-5}{2} \right)^2 &= r^2\\
\\
r^2 &= \frac{17}{2} = 8.5
\end{aligned}
\end{equation}
$

Thus, the equation of the circle is...
$\displaystyle \left( x - \frac{1}{2} \right)^2 + \left( y - \frac{11}{2} \right)^2 = \frac{17}{2}$