## Precalculus (10th Edition)

a) The center is $(h,k)=(3,0)$ and the radius is $r=2$ b) The graph is attached below. c) There are two $x$-intercepts $(5,0)$ and $(1,0)$ There are no $y$-intercepts.
$(a)$ If we start with the standard form of the equation, we can easily get the center and the radius of the circle by finding the constants $h, k$ and $r$: $(x-h)^2+(y-k)^2=r^2$ Here, $2(x-3)^2+2y^2=8$. Divide both sides by $2$ to obtain: $(x-3)^2+y^2=4$ $(x-3)^2+(y-0)^2=2^2$ Therefore the center is $(h,k)=(3,0)$ and the radius is $r=2$. $(b)$ To graph the circle, plot the center $3, 0)$. Then, with a radius of $2$ units, plot the following points: point $2$ units directly above the center: $(3, 2)$ point $2$ units directly below the center: $(3, -2)$ point $2$ units to the left of the center: $(1, 0)$ point $2$ units to the right of the center: $(5, 0)$ Finally, connect these four points using a smooth curve. Refer to the graph above. $(c)$ The $x$- and $y$-intercepts can be calculated by plugging $0$ into $y$ and $x$, respectively. $x$-intercepts: $(x-3)^2+y^2=4$ $(x-3)^2+0^2=4$ $(x-3)^2=4$ $x-3=\pm2$ $x=3\pm2$ $x_1=5$ $x_2=1$ Therefore there are two $x$-intercepts $(5,0)$ and $(1,0)$ $y$-intercepts: $(x-3)^2+y^2=4$ $(0-3)^2+y^2=4$ $9+y^2=4$ $y^2=-5$ this is not possible, therefore there is no $y$-intercepts.