College Algebra, Chapter 2, 2.2, Section 2.2, Problem 58

Find an equation of the circle with center $(-1,5)$ and passes through $(-4,-6)$


Recall that the general equation for the circle with center at $(h,k)$ and
radius $r$ is..


$
\begin{equation}
\begin{aligned}

(x - h)^2 + (y - k)^2 =& r^2
&& \text{Model}
\\
\\
(x - (-1))^2 + (y - 5)^2 =& r^2
&& \text{Substitute the given}
\\
\\
(x + 1)^2 + (y - 5)^2 =& r^2
&& \text{Simplify}
\\
\\
(-4 + 1)^2 + (-6 - 5)^2 =& r^2
&& \text{Substitute the point $(-4, -6)$ to $x$ and $y$ respectively and solve for $r$}
\\
\\
(-3)^2 + (-11)^2 =& r^2
&&
\\
\\
9 + 121 =& r^2
&&
\\
\\
r^2 =& 130
&&

\end{aligned}
\end{equation}
$


Thus, the equation of the circle is..

$(x + 1)^2 + (y - 5)^2 = 130$