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Circle in a parabola

  1. Aug 17, 2006 #1
    A circle with the radius of 1 unit is inscribed in a parabola with the equation of y=x^2. Find the coordinate of the center of the circle.

    The answer given in the book is (0, 5/4).

    I attempt this question by letting the equation of the circle be x^2 + (y-c)^2 = 1, where c is the y-coordinate of the circle. Then, by substitutuing y=x^2 into that equation, I obtain another equation. From the new equation, I try to use the formula b^2 - 4ac = 0. Then, I got the value of c, which is 5/4.

    However, what I do not understand is this: The equation b^2 - 4ac = 0 means that 2 function intercept at only 1 point. But in this case, the circle touches the parabola at 2 points.

    Do I get the answer c=5/4 by chance? Or is there any other method to solve this problem? Maybe the circle does not intercept the parabola at only 2 points? (I say this because I find that at the points of interception, it seems that the circle and the parabola do not share a same tangent line).
  2. jcsd
  3. Aug 17, 2006 #2
    The resulting equation (when you substitute) is quartic: x^4+(1-2c)x^2+c^2=1
    The reason you have two intersection points is because the solution to the quadratic formula in this case is actually x^2

    x^2 = (-b +- sqrt(b^2-4ac))/(2a)

    Even though the discriminant, being zero, causes the square root to drop out of the quadratic forumla (giving just one value on the right) you still have to take the square root of both sides to solve for x here, which gives two values of x in this case.

    Last edited: Aug 17, 2006
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