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


$\begin{equation} \begin{aligned} (x-2)^2 + (y-2)^2 =& (\sqrt{5})^2 \\ (x-2)^2 + (y-2)^2 =& 5 \end{aligned} \end{equation}$