Answer
(a) $(\pm4,0)$, $(0,\pm4)$, (b) symmetric with respect to the x-axis, y-axis, and the origin. (c) See graph.
Work Step by Step
(a) Given $x^2+y^2=16$, we can find the x-intercept(s) (let y=0) as $(\pm4,0)$, y-intercept(s) (let x=0) as $(0,\pm4)$,
(b) To test symmetry, replace $(x,y)$ with $(x,-y)$ (x-axis symmetry), or $(-x,y)$ (y-axis symmetry), or $(-x,-y)$ (origin symmetry), and we can find the equation is symmetric with respect to the x-axis, y-axis, and the origin.
(c) See graph.
