Answer
Only $(0,3)$ is on the graph of $y^{2}=x^{2}+9$.
Work Step by Step
a) If $(0,3)$ lies on the graph of $y^{2}=x^{2}+9$, then it will satisfy the equation. Taking $x=0$ and $y=3$, we get-
$(3)^{2}=(0)^{2}+9$
$9=9$
Since LHS=RHS therefore $(0,3)$ lies on the graph.
b) If $(3,0)$ lies on the graph of $y^{2}=x^{2}+9$, then it will satisfy the equation. Taking $x=3$ and $y=0$, we get-
$(0)^{2}=(3)^{2}+9$
$0\ne18$
Since LHS$\ne$RHS therefore $(3,0)$ does not lie on the graph.
c) If $(-3,0)$ lies on the graph of $y^{2}=x^{2}+9$, then it will satisfy the equation. Taking $x=-3$ and $y=0$, we get-
$(0)^{2}=(-3)^{2}+9$
$0\ne18$
Since LHS$\ne$RHS therefore $(-3,0)$ does not lie on the graph.
