+186 A circle of radius r is tangent to the graph of y=x^2 at two points. Another circle of the same radius is tangent to the graph of y=x^2 and to the x-axis. The two circles and the parabola all share the same point of tangency. Find r.
It seems there was an error in the previous response where the solution was not complete. I apologize for that. Here is the complete solution:
Let (a, a^2) be the point of tangency shared by the two circles and the parabola. The first circle is tangent to the parabola at this point and has its center on the line y = a^2 + r. The second circle is tangent to the x-axis at the point (a - r, 0) and has its center on the line y = -r.
By distance formula, we know that the distance between the centers of the two circles is 2r, and the distance from each circle's center to the point of tangency is also r. Therefore, we have the following equation:
(2r)^2 = (a - (a^2 + r))^2 + (-r - a^2)^2
Solving for a yields:
a = (2r + 1)/2
Also, the equation of the tangent line to the parabola at (a, a^2) is y = -(x - a)^2 + a^2. Since this is a circle of radius r and center (a, a^2 + r), we can write:
(x - a)^2 + (y - a^2 - r)^2 = r^2
Substituting the expression for a we found earlier and solving for r, we get:
r = 1/2
Therefore, the radius of the circles is 1/2.
+186 After playing around on Desmos for a bit, I'm unsure if I can see if a circle with a radius of 1/2 will fit this circumstance. Are you sure this is the right answer? Also, since a=1, this can't possibly be right, since there are no circles that can go through (1,1) with radius 1/2, or yet even other any radius!
