Because both equations involve $(x-4)$ and $(y+2)$, try defining new variables like $u = x-4$ and $v = y+2$ to rewrite both equations in a simpler form.

After rewriting, substitute the expression for $v$ from the parabola into the circle equation. You will get an equation in one variable; reduce it to a quadratic in $t = u^2$ and think about how many real $u$ values that produces.

You can either type the original equations or use the shifted ones. For a clearer view around the shared center/vertex, enter:

• u^2 + v^2 = 25
• v = u^2

in Desmos, treating u as x and v as y (Desmos will still label them as x and y).

If you prefer to use the original variables, enter:

• (x-4)^2 + (y+2)^2 = 25
• y = (x-4)^2 - 2

Adjust the viewing window so that the point $(4,-2)$ is near the center of the screen and the circle is fully visible.

Use Desmos’s intersection tool (click on the graphs where they cross, or tap the gray dots that appear) to mark all intersection points between the circle and the parabola. Count how many distinct intersection points appear; that count tells you which answer choice to select.

Identify each graph:

• Circle: $(x-4)^2 + (y+2)^2 = 25$ has center $(4,-2)$ and radius $5$.
• Parabola: $y = (x-4)^2 - 2$ has vertex $(4,-2)$ and opens upward.

So both graphs are related to the point $(4,-2)$, which suggests shifting the coordinate system so that point becomes the origin. This will simplify the equations.

Let

• $u = x - 4$
• $v = y + 2$

Then $(x-4)^2 = u^2$ and $y+2 = v$.

Rewrite both equations:

• Circle: $(x-4)^2 + (y+2)^2 = 25$ becomes

• Parabola: $y = (x-4)^2 - 2$ means

so in $(u,v)$ this is

Now we need the intersection points of the circle $u^2 + v^2 = 25$ and the parabola $v = u^2$ in the $(u,v)$-plane.

Substitute $v = u^2$ into the circle equation $u^2 + v^2 = 25$:

This rearranges to

Let $t = u^2$ (so $t \ge 0$). Then the equation becomes a quadratic in $t$:

Solve this quadratic using the quadratic formula:

One of these values of $t$ is positive and the other is negative. Remember that $t = u^2$ cannot be negative, so only the positive value of $t$ corresponds to real intersection points.

We found exactly one positive value of $t = u^2$. A single positive $t$ value gives two different real $u$ values:

• $u = +\sqrt{t}$
• $u = -\sqrt{t}$

Each $u$ value gives a corresponding $v$ value using $v = u^2 = t$, so there are two distinct $(u,v)$ intersection points.

Translating back to $(x,y)$ using $x = u + 4$ and $y = v - 2$ does not change how many intersection points there are, only their locations.

Therefore, the graphs intersect at exactly two points.