the center of a circle is at the origin on a coordinate grid. the vertex of a parabola that opens upward is at (0, 9). if the circle intersects the parabola at the parabola’s vertex, which statement must be true?

the maximum number of solutions is one.

the maximum number of solutions is three.

the circle has a radius equal to 3.

the circle has a radius less than 9.

Question

the center of a circle is at the origin on a coordinate grid. the vertex of a parabola that opens upward is at (0, 9). if the circle intersects the parabola at the parabola’s vertex, which statement must be true?

the maximum number of solutions is one.

the maximum number of solutions is three.

the circle has a radius equal to 3.

the circle has a radius less than 9.

Accepted Answer

# Brief Explanations: 1. Analyze the circle and parabola: The circle is centered at the origin $(0,0)$, and the parabola opens upward with vertex at $(0,9)$. They intersect at the vertex $(0,9)$. 2. For the circle's equation $x^{2}+y^{2}=r^{2}$ (center at origin), substituting the vertex $(0,9)$ into the circle's equation gives $0^{2}+9^{2}=r^{2}$, so $r = 9$ (or $r\geq9$ if there are more intersections, but since it's given they intersect at the vertex, the radius must be at least 9? Wait, no—wait, the parabola opens upward, so its equation is $y=a(x - 0)^{2}+9=ax^{2}+9$ ($a>0$). The circle is $x^{2}+y^{2}=r^{2}$. Substituting $y = ax^{2}+9$ into the circle's equation: $x^{2}+(ax^{2}+9)^{2}=r^{2}$. This is a quartic equation, which can have up to 4 real solutions. But let's check the radius: the distance from the origin (center of the circle) to the vertex $(0,9)$ is 9. Since the circle intersects the parabola at $(0,9)$, the radius must be at least 9? Wait, no—wait, the options: - Option 1: "The maximum number of solutions is one"—a quartic can have up to 4, so no. - Option 2: "The maximum number of solutions is three"—quartic can have 4, so no. - Option 3: "The circle has a radius equal to 3"—distance from origin to $(0,9)$ is 9, so radius 3 would not reach $(0,9)$, so no. - Option 4: Wait, wait, maybe I made a mistake. Wait, the parabola opens upward, vertex at $(0,9)$, so the parabola is above or at $y = 9$ (since it opens upward, vertex is the minimum point). The circle is centered at origin. The intersection is at $(0,9)$, so the radius must be at least 9? But the option says "The circle has a radius less than 9"—no, that can't be. Wait, maybe I misread the options. Wait, the original problem: let's re - check. Wait, maybe the options were miswritten? Wait, no—wait, the parabola opens upward, so its vertex is the minimum point. So the parabola is $y = ax^{2}+9$, $a>0$, so for $x eq0$, $y>9$. The circle is $x^{2}+y^{2}=r^{2}$. At $x = 0$, $y=\pm r$. The intersection at $(0,9)$ means that when $x = 0$, $y = 9$ is on the circle, so $r\geq9$ (if $r = 9$, then at $x = 0$, $y = 9$ is on the circle; if $r>9$, then there might be other intersections). But the options: Wait, the options given: - "The maximum number of solutions is one": No, because the system of equations (circle and parabola) is a quartic, which can have up to 4 solutions. - "The maximum number of solutions is three": A quartic equation can have 0, 1, 2, 3, or 4 real roots. But since the parabola opens upward and the circle is centered at the origin, let's think about the graphs. The parabola has a minimum at $(0,9)$, and the circle is centered at $(0,0)$. If the radius is 9, the circle passes through $(0,9)$. For $x eq0$, the parabola is above $y = 9$, and the circle at $x eq0$ has $y=\sqrt{r^{2}-x^{2}}$. If $r = 9$, then for $x eq0$, $y=\sqrt{81 - x^{2}}<9$, but the parabola at $x eq0$ has $y>9$, so no other intersections. If $r>9$, then for some $x$, $\sqrt{r^{2}-x^{2}}>9$, so we can have two more intersections (since the equation $x^{2}+(ax^{2}+9)^{2}=r^{2}$ is even in $x$, so solutions come in pairs $(x,y)$ and $(-x,y)$). So the maximum number of solutions can be 3? No, wait, if $r>9$, let's take $a = 1$, $r = 10$. Then the equation is $x^{2}+(x^{2}+9)^{2}=100$, $x^{2}+x^{4}+18x^{2}+81 = 100$, $x^{4}+19x^{2}-19 = 0$. Let $u=x^{2}$, then $u^{2}+19u - 19=0$, which has one positive root ($u=\frac{-19+\sqrt{361 + 76}}{2}=\frac{-19+\sqrt{437}}{2}\approx\frac{-19 + 20.9}{2}\approx0.95$), so $x=\pm\sqrt{0.95}$, so two more solutions, plus $(0,9)$, total three? Wait, no, $x = 0$ gives $y = 9$, and $x=\pm\sqrt{0.95}$ gives two points, so total three solutions? Wait, but a quartic can have up to 4. Wait, maybe my analysis is wrong. Wait, let's go back to the options. The key is the radius: the distance from the origin to $(0,9)$ is 9. Since the circle intersects the parabola at $(0,9)$, the radius must be at least 9? But the option "The circle has a radius less than 9"—no, that can't be. Wait, maybe the original problem has a typo, or I misread the direction of the parabola. If the parabola opens downward, vertex at $(0,9)$, then $y=-ax^{2}+9$, $a>0$, and then the circle with radius 9 would intersect at $(0,9)$, and for $x eq0$, $y=-ax^{2}+9<9$, and the circle at $x eq0$ has $y=\sqrt{81 - x^{2}}<9$, so no other intersections. But the problem says "opens upward". Wait, maybe the correct answer is "The maximum number of solutions is three"—but no, let's check the options again. Wait, the first option: "The maximum number of solutions is one"—if the radius is 9, then only $(0,9)$ is the intersection, so maximum one? No, because if the radius is greater than 9, we can have three solutions (as above). Wait, no, a quartic equation can have up to 4 real roots, but in this case, due to the symmetry (parabola is symmetric about y - axis, circle is symmetric about y - axis), the solutions are symmetric in x. So if $x$ is a solution, $-x$ is a solution. So the number of solutions is either 1 (when $x = 0$ is the only solution) or 3 (when $x = 0$ and $x=\pm k$, $k eq0$) or 4 (but in this case, since the parabola opens upward, $y=ax^{2}+9\geq9$, and the circle $x^{2}+y^{2}=r^{2}$, so $y\leq r$ (since $x^{2}\geq0$). So for $y>9$, we have $ax^{2}+9\leq r$, $x^{2}\leq\frac{r - 9}{a}$. So if $r>9$, there are two solutions for $x$ (positive and negative) and $x = 0$, total three. If $r = 9$, only $x = 0$. So the maximum number of solutions is three. So the correct option is "The maximum number of solutions is three". # Answer: The maximum number of solutions is three.

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