Because the second equation already gives $y$ in terms of $x$, try substituting $y = 2x - x^2$ into the first equation. This will give you a single equation in $x$.

After substitution, you will get a polynomial equation in $x$. Factor out any simple roots you can find (like $x = -1$, $0$, $1$, $2$, etc.), and then think about how many real roots remain.

Each real solution for $x$ corresponds to one point on the parabola, and you already ensured it lies on the circle. How does the number of real $x$-solutions relate to the number of ordered pairs $(x,y)$ that satisfy the system?

In Desmos, type y = 2x - x^2 for the parabola and (x - 3)^2 + y^2 = 25 for the circle. Desmos will automatically plot both graphs.

Use the tap/click feature on the points where the parabola and circle cross. Desmos will highlight each intersection and display its coordinates.

Count how many distinct intersection points appear between the two graphs. That count is the number of ordered pairs $(x,y)$ that satisfy the system.

Rewrite the equations in a more recognizable form:

• $(x - 3)^2 + y^2 = 25$ is a circle with center $(3,0)$ and radius $5$.
• $y = 2x - x^2$ is a downward-opening parabola with vertex at $(1,1)$.

The question is asking how many points these two graphs have in common (how many intersection points).

At any intersection point, the same $(x,y)$ must satisfy both equations.

Since $y = 2x - x^2$, substitute this expression for $y$ into the circle equation:

Now the equation involves only $x$, so solving it will give the $x$-coordinates of intersection points.

Expand each term:

• $(x - 3)^2 = x^2 - 6x + 9$.
• $(2x - x^2)^2 = x^4 - 4x^3 + 4x^2$.

Substitute and set equal to $25$:

Combine like terms and move $25$ to the left:

Now we need to find how many real solutions this quartic equation has.

Look for an easy factor by testing small integer values. Try $x = -1$:

so $x = -1$ is a root, and $(x + 1)$ is a factor.

Divide $x^4 - 4x^3 + 5x^2 - 6x - 16$ by $(x + 1)$ to get

We already have one real solution $x = -1$. Now consider the cubic $x^3 - 5x^2 + 10x - 16$:

• At $x = 2$: $2^3 - 5(2)^2 + 10(2) - 16 = 8 - 20 + 20 - 16 = -8$ (negative).
• At $x = 4$: $4^3 - 5(4)^2 + 10(4) - 16 = 64 - 80 + 40 - 16 = 8$ (positive).

Since the cubic is continuous and changes sign between $x = 2$ and $x = 4$, it has one real root in that interval. (A cubic cannot have more than three real roots total, and a sign change on exactly one interval shows at least one and suggests isolating that one.)

Thus the quartic has exactly two real $x$-values: $x = -1$ and one value between $2$ and $4$. Each $x$ gives exactly one $y$ from $y = 2x - x^2$, so there are exactly 2 ordered pairs $(x,y)$ that satisfy the system. The correct choice is B) There are exactly 2 solutions.