Crossing Curves: Finding the Intersections of Parabolas and Circles

By Mary Jane Sterling

When a parabola and circle intersect, the possibilities for their meeting are many and varied. The two curves can intersect in as many as four different points, or maybe three, or just two or even just one point.

Keep your options open and be alert for as many common solutions as possible (right — up to four). And, yes, the system may have no solution at all. The curves may miss each other completely.

For these problems, you usually turn to substitution. However, you don’t have to set one of the equations equal to x or y by itself. You may solve an equation for 4x or (y – 3)2 or some other term that appears in the other equation. As long as the terms match, you can swap one value for the other.

Sample questions

  1. Find the common solutions of the circle (x – 2)2 + (y – 2)2 = 4 and the parabola 2y = x2 – 4x + 4.

    (0, 2), (2, 0), (4, 2). Rewrite the equation of the parabola as 2y = (x – 2)2. Next, replace the (x– 2)2 term in the first equation with 2y and simplify: 2y + (y – 2)2 = 4; 2y + y2 – 4y + 4 = 4; y2 – 2y = 0.

    Factor the terms on the left to get y(y – 2) = 0. So, y = 0 or y = 2. Letting y = 0 in the equation of the parabola, you get 2(0) = x2 – 4x + 4, or 0 = (x – 2)2. So, when y = 0, x = 2.

    Next, let y = 2 in the parabola equation. You get 2(2) = x2 – 4x + 4; 4 = x2 – 4x + 4; 0 = x2 – 4x. Factoring, 0 = x(x – 4), so x = 0 or x = 4. When y = 2, x = 0 or 4.

  2. Find the common solutions of x2 + y2 = 100 and y2 + 6x = 100.

    (0, 10), (0, –10), (6, 8), (6, –8). Solving the second equation for y2, you get y2 = 100 – 6x. Replace the y2 in the first equation with its equivalent to get x2 + 100 – 6x = 100.

    Simplifying and factoring, the equation becomes x2 – 6x = x(x – 6) = 0. So x = 0 or x = 6. Replacing x with 0 in the equation of the parabola, y2 = 100; y = +/–10. Replacing x with 6 in the equation of the parabola, y2 + 36 = 100; y2 = 64; y = +/–8.

Practice questions

  1. Find the common solutions of x2 + y2 = 25 and x2 + 4y = 25.

  2. Find the common solutions of x2 + y2 = 9 and 5x2 – 6y = 18.

Following are answers to the practice questions:

  1. The answer is (5, 0), (–5, 0), (3, 4), (–3, 4).

    Solve the second equation for x2 (you get x2 = 25 – 4y) and replace the x2 in the first equation with its equivalent. The new equation reads 25 – 4y + y2 = 25. Simplifying, you get y2 – 4y = 0.

    This equation factors into y(y – 4) = 0. The two solutions are y = 0 and y = 4. Go back to the second equation, the equation of the parabola, because it has only one squared term (it has lower exponents, so choosing this equation lets you avoid extraneous solutions).

    Replace the y in that equation with 0 to get x2 + 4(0) = 25; x2 = 25. That equation has two solutions: x = 5 or x = –5. Now, going back and replacing the y with 4 in the equation of the parabola, x2 + 4(4) = 25; x2 + 16 = 25; x2 = 9.

    This equation also has two solutions: x = 3 or x = –3. Pairing up the y’s and their respective x’s, you get the four different solutions. The circle and parabola intersect in four distinct points.

  2. The answer is

    image0.jpg
    image1.jpg

    (0, –3).

    Eliminate the x terms: Multiply the terms of the first equation by –5 (which gives you –5x2 – 5y2= –45) and add the two equations together. The resulting equation is –5y2 – 6y = –27.

    Rewrite the equation, setting it equal to 0, and factor. You get 0 = 5y2 + 6y – 27 = (5y – 9)(y + 3). Use the multiplication property of zero to solve for the two solutions of this equation. Replacing the y in the second equation (the equation of the parabola) with

    image2.jpg

    you get

    image3.jpg

    Then divide both sides of the equation by 5 and take the square root of each side:

    image4.jpg

    To find the other solution, let the y in the equation of the parabola be equal to –3. You get 5x2 – 6(–3) = 18; 5x2 + 18 = 18; 5x2 = 0. So, x = 0. The circle and parabola intersect or touch in three distinct points.