Precalculus, Chapter 1, 1.4, Section 1.4, Problem 40

Determine the standard form of the equation of a circle with endpoints of a diameter at $(4,3)$ and $(0,1)$.

Using the Distance Formula to find the diameter of the circle,


$
\begin{equation}
\begin{aligned}

D =& \sqrt{(3-1)^2 + (4-0)^2}
\\
D =& \sqrt{4 + 16}
\\
D =& \sqrt{20}

\end{aligned}
\end{equation}
$


We know that $\displaystyle r = \frac{D}{2}$, where $D$ is the diameter and $r$ is the radius. So we have


$
\begin{equation}
\begin{aligned}

r =& \frac{2 \sqrt{5}}{5}
\\
r =& \sqrt{5}

\end{aligned}
\end{equation}
$


To find the center of the circle, we use the Midpoint Formula


$
\begin{equation}
\begin{aligned}

M =& \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\\
\\
=& \left( \frac{4+0}{2}, \frac{3+1}{2} \right)
\\
\\
=& (2,2)

\end{aligned}
\end{equation}
$


The standard from of the equation of a circle with center $(2,2)$ and radius $\sqrt{5}$ is


$
\begin{equation}
\begin{aligned}

(x-2)^2 + (y-2)^2 =& (\sqrt{5})^2
\\
(x-2)^2 + (y-2)^2 =& 5

\end{aligned}
\end{equation}
$