Mathematica and Problem to solve...DIFFICULT!?

Question

The figure shows a circle with radius 1 inscribed in the parabola y=x^2 . Find the center of the circle.

Please give me step by step instructions if you can!
Thanks

Answer 1

Follow these steps:

Solve the equation for the intersection points of a circle and a parabola, for a circle of radius 1 and with center at (0,h):

h - √(1-x²) = x²

Solve for x, and get 4 roots. Take two positive or two negative roots and assume that they are equal (double roots). This is the same as making the circle tangent to the parabola:

√(-(1/2)+h - (1/2)√(5-4h)) = √(-(1/2)+h + (1/2)√(5-4h))

Solve for h, and find that h = 5/4, which is your answer. You can verify by plotting circle and parabola and see that they are tangent at 2 points.

Addendum: The high school way to solve this is to find the equation of the line perpendicular to the tangent of the parabola at (z, z²), which is -(1/2z)x + z² + 1/2. At z = 0, y = z² + 1/2, so we have the end points of the radius of the circle (0, z² + 1/2) and (z, z²). Using Pythagorean's theorem and that the radius = 1, we find that z = (1/2)√3, and from this and using y = z² + 1/2, we have y = 5/4 as worked out using the other way.

Answer 2

if the circle touches to origin C(0,1)

if not solve them together
find the intersection points

y=x^2
x^2+y^2=1

Answer 3

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