After substitution and expanding, you will get an equation that includes $x^4$. Notice that it only contains even powers of $x$, so you can set $t = x^2$ to turn it into a quadratic equation in $t$.

Once you solve for $t = x^2$, remember each positive value of $x^2$ leads to two possible $x$ values, $x = \sqrt{t}$ and $x = -\sqrt{t}$. For each $x$, use $y = x^2 - 4$ to find the matching $y$, then count all distinct $(x,y)$ pairs.

In Desmos, type x^2 + y^2 = 10 to graph the circle. Make sure to include =, not y =, so Desmos treats it as a relation and draws the full circle.

In a new line, type y = x^2 - 4 to graph the parabola that opens upward and is shifted down 4 units.

Look for the points where the parabola intersects the circle. Tap each intersection point to see its coordinates, and count how many distinct intersection points there are; that count is the number of real ordered pairs $(x,y)$ that satisfy the system.

Use the second equation, $y = x^2 - 4$, and substitute it into the first equation $x^2 + y^2 = 10$.

That gives

Now expand and simplify $(x^2 - 4)^2$:

Substitute this into the equation:

This is a quartic in $x$, but it only involves $x^2$. Let $t = x^2$. Then the equation becomes

Factor the quadratic in $t$:

So

Recall $t = x^2$, so

That gives

• From $x^2 = 1$: $x = 1$ or $x = -1$.
• From $x^2 = 6$: $x = \sqrt{6}$ or $x = -\sqrt{6}$.

Now find $y$ using $y = x^2 - 4$:

• If $x^2 = 1$, then $y = 1 - 4 = -3$.
• If $x^2 = 6$, then $y = 6 - 4 = 2$.

So the solution points are

• $(1, -3)$ and $(-1, -3)$ from $x^2 = 1$,
• $(\sqrt{6}, 2)$ and $(-\sqrt{6}, 2)$ from $x^2 = 6$.

We have found four distinct real ordered pairs that satisfy both equations:

• $(1, -3)$
• $(-1, -3)$
• $(\sqrt{6}, 2)$
• $(-\sqrt{6}, 2)$

Therefore, the number of distinct real ordered pairs $(x,y)$ that satisfy the system is $4$, which corresponds to choice D.