Precalculus, Chapter 1, Review Exercises, Section Review Exercises, Problem 26

Determine the center and radius of the circle $2x^2 + 2y^2 - 4x = 0$. Graph the circle. Find the intercepts, if any.

We first simplify the equation


$
\begin{equation}
\begin{aligned}

2x^2 + 2y^2 - 4x =& 0
&& \text{Given equation}
\\
2 \left( x^2 + y^2 - 2x \right) =& 0
&& \text{Factor out } 2
\\
x^2 + y^2 - 2x =& 0
&& \text{Simplified form}
\\
(x^2 - 2x) + y^2 =& 0
&& \text{Group the equation in terms of $x$ and $y$. And put the consistent on the right side of the equation}
\\
(x^2 - 2x + 1) + y^2 =& 0 + 1
&& \text{Complete the square: add } \left( \frac{2}{2} \right)^2 = 1
\\
(x -1)^2 + y^2 =& 1
&& \text{Factor}

\end{aligned}
\end{equation}
$


We recognize this equation as the standard form of the equation of a circle with $r=1$ and center $(1,0)$







To find the $x$-intercepts, we let $y = 0$. Then


$
\begin{equation}
\begin{aligned}

(x-1)^2 + y^2 =& 1
&&
\\
(x - 1)^2 + 0 =& 1
&& y = 0
\\
(x-1)^2 =& 1
&& \text{Simplify}
\\
x - 1 =& \pm 1
&& \text{Solve for } x
\\
x =& 1 \pm 1
&&

\end{aligned}
\end{equation}
$


The $x$-intercepts are $2$ and .

To find the $y$-intercepts, we let $x = 0$. Then


$
\begin{equation}
\begin{aligned}

(x-1)^2 + y^2 =& 1
&&
\\
(0-1)^2 + y^2 =& 1
&& x = 0
\\
1 + y^2 =& 1
&& \text{Simplify}
\\
y^2 =& 0
&& \text{Solve for } y
\\
y =& 0
&&

\end{aligned}
\end{equation}
$


The $y$-intercept is .