College Algebra, Chapter 2, 2.2, Section 2.2, Problem 64

Find the equation of the circle shown in the figure.

*Refer to the figure in the book*

By observation, the center of the circle is at $(-1, 1)$ and it passes through point $(2, 0)$. Recall that the general equation for the circle with
circle $(h,k)$ and radius $r$ is..



$
\begin{equation}
\begin{aligned}

(x - h)^2 + (y - k)^2 =& r^2
&& \text{Model}
\\
\\
(x - (-1))^2 + (y - 1)^2 =& r^2
&& \text{Substitute the value of the center}
\\
\\
(x + 1)^2 + (y - 1)^2 =& r^2
&& \text{Simplify}

\end{aligned}
\end{equation}
$


Since the circle passes through the point $(2,0)$, we can say that the point is a solution for the equation.


$
\begin{equation}
\begin{aligned}

(2 + 1)^2 + (0 - 1)^2 =& r^2
\\
\\
(3)^2 + (-1)^2 =& r^2
\\
\\
9 + 1 =& r^2
\\
\\
10 =& r^2

\end{aligned}
\end{equation}
$


Thus, the equation of the circle is..

$(x + 1)^2 + (y - 1)^2 = 10$

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