Precalculus, Chapter 1, 1.4, Section 1.4, Problem 10

Determine the center and radius of the given circle. Write the standard form of the equation.







Using the distance formula, to solve for the diameter, but $\displaystyle r = \frac{d}{2}$. So we have


$
\begin{equation}
\begin{aligned}

r =& \frac{\sqrt{(1-3)^2 + (0-2)^2}}{2}
\\
\\
r =& \frac{\sqrt{4+4}}{2}
\\
\\
r =& \frac{\sqrt{8}}{2}
\\
\\
r =& \frac{2 \sqrt{2}}{2}
\\
\\
r =& \sqrt{2}

\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{0+2}{2}, \frac{1+3}{2} \right)
\\
\\
=& \left( \frac{2}{2} , \frac{4}{2} \right)
\\
\\
=& (1,2)

\end{aligned}
\end{equation}
$


So the center of the circl is $(1,2)$.

The equation of the circle in standard form is



$
\begin{equation}
\begin{aligned}

(x-1)^2 + (y-2)^2 =& (\sqrt{2})^2
\\
(x-1)^2 + (y-2)^2 =& 2

\end{aligned}
\end{equation}
$