Answer
(a) $(-2,2)$, $3$. (b) see graph. (c) $(-2\pm\sqrt 5,0)$, $(0,2\pm\sqrt {5})$
Work Step by Step
(a) Given $x^2+y^2+4x-4y-1=0$, form squares to get $(x+2)^2+(y-2)^2=9$. We can identify its center $(h,k)=(-2,2)$ and radius $r=3$.
(b) see graph.
(c) To find the x-intercepts, letting $y=0$, we have $x^2+4x-1=0$ which gives $(-2\pm\sqrt 5,0)$. To find the y-intercepts, letting $x=0$, we have $y^2-4y-1=0$ which gives $(0,2\pm\sqrt {5})$
