After substituting $y = -2$, remember that $(-2)^2$ is positive. What does the equation become after you simplify?

Once you have an equation involving $x^2$, isolate $x^2$ and then take square roots. Don’t forget that there can be both a positive and a negative square root.

You should get two possible values of $x$. Check which of those values are actually listed among the answer choices.

Type x^2 + y^2 = 13 on one line and y = -2 on another line in Desmos. This will graph a circle and a horizontal line.

Look for the points where the line $y=-2$ intersects the circle $x^2 + y^2 = 13$. Click on each intersection point to see its coordinates.

Note the $x$-coordinates of the intersection points. These are the possible values of $x$ for the system. Compare these $x$-values with the answer choices and select the matching one.

We are told $y=-2$. Substitute this into the first equation $x^2 + y^2 = 13$:

Now simplify $(-2)^2$.

Compute $(-2)^2$:

So the equation becomes

Subtract 4 from both sides:

To solve $x^2 = 9$, take the square root of both sides. Remember that both positive and negative roots are possible:

These are the two $x$-values that work with $y = -2$ in the original system.

From the solutions $x=3$ and $x=-3$, check which one appears in the choices A–D.

Only $-3$ is listed (choice D), so the possible value of $x$ from the given options is $-3$.