Answer
$(-3,0), (3,0)$
Work Step by Step
We are given the curves:
$\begin{cases}
x^2-y^2=9\\
x^2+y^2=9
\end{cases}$
The equation $x^2-y^2=9$ represents a horizontal hyperbola centred in origin. The equation $x^2+y^2=9$ represents a circle centered in origin and having radius 3.
We graph both curves.
From the graph we find the intersection points:
$(-3,0)$ and $(3,0)$
We check if both points check the equations:
$(-3,0)$
$x^2-y^2=9$
$(-3)^2-0^2\stackrel{?}{=}9$
$9=9\checkmark$
$x^2+y^2=9$
$(-3)^2+0^2\stackrel{?}{=}9$
$9=9\checkmark$
$(3,0)$
$x^2-y^2=9$
$3^2-0^2\stackrel{?}{=}9$
$9=9\checkmark$
$x^2+y^2=9$
$3^2+0^2\stackrel{?}{=}9$
$9=9\checkmark$
So the system's solutions are:
$(-3,0), (3,0)$
