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 .