Answer
Please see the graph.
Work Step by Step
Orange line: $x^2+y^2 < 4$
Red line: $x^2-y^2 \leq 1$
$x^2-y^2 \leq 1$
This is the graph of a hyperbola, so we have three regions to test for shading. These regions (x-values only) are $(-∞,-1)$, $(-1,1)$, and $(1,∞)$.
We pick the points $(-3,0)$, $(0,0)$, and $(3,0)$.
$(-3,0)$
$x^2-y^2 \leq 1$
$(-3)^2-0^2 \leq 1$
$9 - 0 <1$
$9 < 1$ (false, so we don't shade this area)
$(0,0)$
$x^2-y^2 \leq 1$
$0^2-0^2 \leq 1$
$0 - 0 \leq 1$
$0 \leq 1$ (true, so we shade this area)
$(3,0)$
$x^2-y^2 \leq 1$
$(3)^2-0^2 \leq 1$
$9 - 0 \leq 1$
$9 \leq 1$ (false, so we don't shade this area)
$x^2+y^2 < 4$
This inequality is for a circle with radius 2. We pick the point $(0,0)$ to determine whether to shade inside or outside the circle.
$x^2+y^2 < 4$
$0^2+0^2 < 4$
$0 + 0 < 4$
$0 < 4$ (true, so we shade inside the circle)
The overlap of the two graphs is the solution set for the inequalities.
