Question 90·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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A parabola and a circle are represented by the following system, where $r$ is a positive constant:

\begin{aligned} y &= (x-3)^2+1 \\[1.4em] (x-3)^2+(y-6)^2 &= r^2 \end{aligned}

The system has exactly three distinct real solutions $(x,y)$ . Which choice is the value of $r$ ?

A parabola and circle with the same line of symmetry have intersections that come in mirrored pairs. An odd number of distinct intersections means one must lie on that line, so find that point on the parabola and measure its distance from the circle’s center. Then graph both equations in Desmos to confirm the count; a nearby radius can produce a different number of intersections.

Hints

Hint 1
Both equations contain $(x-3)^2$ . Points equally far to the left and right of $x=3$ give the same square, so an intersection away from that line has a mirror partner on the other side.
Hint 2
The question asks for an odd number of distinct solutions. Mirrored pairs account for an even number, so where must the remaining intersection lie? Find the parabola’s point on that line.
Hint 3
A circle’s radius is the distance from its center to a point on its edge. Read the center from the circle equation, then find the vertical distance to the required point on the parabola.

Step-by-step

Use symmetry, then confirm in Desmos
Step 1
Pair the off-axis intersections
For any distance $d$ , $x=3-d$ and $x=3+d$ give the same $(x-3)^2=d^2$ in both equations. So an intersection on one side of $x=3$ has a mirror partner on the other. The shared line of symmetry is $x=3$ .
Step 2
Locate the unpaired intersection
The system has exactly three distinct solutions. Points off $x=3$ come in pairs, so an odd intersection count requires a shared point on the symmetry line. One solution must have $x=3$ .
Step 3
Find that point on the parabola
In $y=(x-3)^2+1$ , the square is zero when $x=3$ , giving $y=1$ . The parabola’s vertex, its lowest point, is $(3,1)$ . This must be the intersection on the symmetry line.
Step 4
Measure from the circle’s center
In a circle equation $(x-h)^2+(y-k)^2=r^2$ , the center is $(h,k)$ and the radius $r$ is the distance to its edge. This circle’s center is $(3,6)$ . The required point $(3,1)$ lies straight below it, so $r=|6-1|$ .
Step 5
Confirm the three solutions
Type $r=|6-1|$ and both original equations into Desmos. It prints $r=5$ . Click the three shared points to see $(0,10)$ , $(3,1)$ , and $(6,10)$ . So this radius gives exactly three distinct solutions, and $r=5$ . Choice A.

Question 90·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

Hints

Step-by-step

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