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 parabolas 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.

Explanation:

Step1: Analyze the circle and parabola

The circle is centered at the origin \((0,0)\), and the parabola has its vertex at \((0,9)\) and opens upward. The intersection point is the vertex \((0,9)\). The distance from the center of the circle (origin) to the intersection point \((0,9)\) is the radius of the circle (since the intersection is on the circle). The distance formula between \((0,0)\) and \((0,9)\) is \(d = \sqrt{(0 - 0)^2+(9 - 0)^2}=9\). But if the circle only intersects the parabola at the vertex, the radius must be such that the circle just touches or is smaller than the distance to the vertex (because the parabola opens upward, for \(x
eq0\), the \(y\)-value of the parabola will be less than 9? Wait, no, the parabola opens upward, so its equation is \(y=a(x - 0)^2+9=ax^{2}+9\) with \(a>0\). For \(x
eq0\), \(y>9\). The circle's equation is \(x^{2}+y^{2}=r^{2}\). At the intersection point \((0,9)\), plugging into the circle's equation: \(0 + 9^{2}=r^{2}\), so \(r = 9\) if it passes through \((0,9)\). But if the circle only intersects the parabola at \((0,9)\), then for \(x
eq0\), the points on the parabola \((x,ax^{2}+9)\) must not lie on the circle. Let's check the circle's equation for a point \((x,ax^{2}+9)\) on the parabola: \(x^{2}+(ax^{2}+9)^{2}=r^{2}\). At \(x = 0\), this is \(81=r^{2}\), so \(r = 9\). For \(x
eq0\), \(x^{2}+(ax^{2}+9)^{2}>x^{2}+81\) (since \(a>0\), so \((ax^{2}+9)^{2}>81\) when \(x
eq0\)). So if the circle has radius \(r\), then for the circle to only intersect the parabola at \((0,9)\), we need that for \(x
eq0\), \(x^{2}+(ax^{2}+9)^{2}>r^{2}\). But at \(x = 0\), \(r^{2}=81\). So if \(r<9\), then \(x^{2}+(ax^{2}+9)^{2}>81>r^{2}\) for all \(x
eq0\), so the only intersection is at \((0,9)\). Now let's analyze the options:

• Option 1: The maximum number of solutions. A circle and a parabola can intersect at 0, 1, 2, 3, or 4 points. So maximum number can be 4, so this is false.
• Option 2: Maximum number of solutions is three. As above, it can be 4, so false.
• Option 3: The circle has radius 3. We saw the radius at the intersection point is 9, so if radius is 3, the circle would not reach \((0,9)\), so false.
• Option 4: The circle has radius less than 9. If \(r<9\), then the only intersection is at \((0,9)\) (since for \(x

eq0\), the parabola's \(y\)-value is greater than 9, and the circle's \(y\)-value at a given \(x\) is \(\sqrt{r^{2}-x^{2}}\), which for \(r<9\) and \(x
eq0\), \(\sqrt{r^{2}-x^{2}}<9\), but the parabola's \(y\) at \(x
eq0\) is \(ax^{2}+9>9\), so the circle (with \(r<9\)) will have \(y=\sqrt{r^{2}-x^{2}}<9\) and the parabola has \(y>9\) for \(x
eq0\), so no intersection there. So the only intersection is at \((0,9)\). So this statement must be true.

Answer:

The circle has a radius less than 9. (The option corresponding to this statement, assuming the options are labeled as: The option with "The circle has a radius less than 9" as its text)