Parabola and a Circle

The unknowns are the circle radius $r$ and the and the $x$ coordinate of the parabola tangent point. Define some tangential and normal unit vectors to the parabola at the tangent point:

$T_x = \frac{1}{\sqrt{1 + 4 x^2}} \\ T_y = \frac{2x}{\sqrt{1 + 4 x^2}} \\ N_x = T_y \\ N_y = -T_x$

Then the equations to be satisfied are:

$x + r N_x + r = \frac{7}{6} \\ x^2 + r N_y - r = 0$

Solving yields $(x,r) = \Big( \frac{2}{3}, \frac{5}{18} \Big)$

Let $r$ be the radius of the circle and let $(p, p^2)$ be the point of tangency between the parabola and the circle. Since the slope of any point on the parabola $y = x^2$ is $y' = 2x$ , the slope of the tangent line is $2p$ , and since the hypotenuse of the given triangle is perpendicular to the tangent line, the sides of the triangle are $t$ and $2pt$ for some variable $t$ .

From the horizontal segments, $p + 2pt + r = \frac{7}{6}$ , and from the vertical segments, $r + t = p^2$ , and from Pythagorean's Theorem, $t^2 + (2pt)^2 = r^2$ .

These three equations solve to positive solutions of $p = \frac{2}{3}$ , $t = \frac{1}{6}$ , and $r = \frac{5}{18}$ , so $a = 5$ , $b = 18$ , and $a + b = \boxed{23}$ .

There are actually exist 4 solutions for this problem. Check the following graph !! (by "Ragai Awad" one of my students)