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Chapter 6: Problem 77

A circle and a parabola can have \(0,1,2,3,\) or 4 points of intersection. Sketch the circle \(x^{2}+y^{2}=4 .\) Discuss how this circle could intersect a parabola with an equation of the form \(y=x^{2}+C .\) Then find the values of \(C\) for each of the five cases described below. Use a graphing utility to verify your results. (a) No points of intersection (b) One point of intersection (c) Two points of intersection (d) Three points of intersection (e) Four points of intersection

Short Answer

Expert verified
To solve the problem, we need to understand the graphs of the given circle and parabola then analyze how they intersect. (a) No intersection when \(C < -4\) or \(C > 4\). (b) One point of intersection when \(C = -4\) or \(C = 4\). (c) Two points of intersection when \(-4 < C < 4\). (d) It's not possible for a parabola to intersect a circle at three points. (e) It's also not possible for a parabola to intersect a circle at four points.

Step by step solution

01

Understanding the graphs

We need to get an understanding of the graphs of a circle with equation \(x^2 + y^2 = 4\) and a parabola \(y = x^2 + C\). The circle will be centered at the origin (0,0) with a radius of 2. The parabola will have a vertex at (0,C) and is an upward-opening parabola.
02

Case (a) No points of intersection

For no points of intersection, the parabola must be either entirely above or below the circle. Since the circle's radius is 2, the parabola is entirely above the circle when \(C > 4\), and is entirely below when \(C < -4\). For this case, the value of \(C\) must be greater than 4 or less than -4.
03

Case (b) One point of intersection

For there to be exactly one point of intersection, the vertex of the parabola must lie on the circle. Therefore, setting \(C = 4\) or \(C = -4\) will create exactly one intersection point.
04

Case (c) Two points of intersection

For two points of intersection, the parabola must intersect the circle at two points. Solving the system of equations \((x^{2} + y^{2} = 4)\) and \((y = x^{2} + C)\) yields the quadratic equation \((x^{4} - 4x^{2} + 4 - C = 0)\). We need the discriminant of this equation to be greater than zero, giving two real solutions. This leaves us with -4 < C < 4.
05

Case (d) Three points of intersection

A parabola cannot intersect a circle at exactly three points. So there is no possible value of for this case.
06

Case (e) Four points of intersection

The situation where a parabola intersects a circle at exactly four points is not possible. Hence, this case has no value of \(\).
07

Verification

Use a graphing utility to sketch the graphs of the circle and the parabolas for different values of \(C\). This is to confirm that the number of intersection points correspond with the computed values for \(C\).

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