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..

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