A tight fit ....

Ujjwal From the figure:

Perpendicular TM dropped from point of tangency T on the x-axis

A, B = X & Y intercepts of tangent at T

AO = OM . . . property of parabola

Hence AB = BT

In triangles BTC2 and ATC1

BT = r/tan(beta) and AT = r/tan(alfa)

Hence 2 tan(alfa) = tan(beta) . . . . . . (I)

Angles 2(alfa) + 2(beta) = 90 . . . complimentary angles in right triangle AOB

Hence, alfa = 45 - beta . . . . . (II)

Substitute in (I) to get the quadratic: t^2 + 3t - 2 = 0 . . . t = tan(beta)

Solve to get beta = 29.32° and alfa = 15.68°

Then find x coordinate of T or OM = 0.6727

Then BT = OM/cos(2 alfa) = 0.7879

and finally r = BT tan(beta) = 0.4424

I'll post a full solution later when I have time, but for now, here is a diagram of the winning scenario.

Edit: O.k., here is an outline of my approach....

Let $P(p,\sqrt{p})$ be the point on $y = \sqrt{x}$ to which both of the radius $r$ circles are tangent. Next, let $A(a,r)$ be the center of the circle that is tangent to both $y = \sqrt{x}$ and the positive $x$ -axis and $B(r,b)$ be the center of the circle tangent to both $y = \sqrt{x}$ and the positive $y$ -axis.

Now the slope of the tangent line to $y = \sqrt{x}$ at $P$ is $\frac{1}{2\sqrt{p}}$ , so the slope of the line through $A$ and $B$ is $-2\sqrt{p}$ . Thus the equation of this normal line is

$y - \sqrt{p} = -2\sqrt{p}(x - p)$ .

Now since $A(a,r)$ lies on this line we have that

$\sqrt{p} - r = 2\sqrt{p}(a - p)$ , (I).

Next, since $P$ is the midpoint of $AB$ we have that

$\dfrac{a + r}{2} = p \Longrightarrow a = 2p - r$ .

Substituting this into equation (I) gives us

$\sqrt{p} - r = 2\sqrt{p}(p - r)$ , (II).

Now let $Q$ be the point of intersection of the vertical line through $B$ and the horizontal line through $A$ . Then $\Delta OAB$ is a right triangle with hypotenuse $AB$ of length $2r$ , base $QA$ of length $a - r = 2p - 2r = 2(p - r)$ and vertical side $QB$ having length $2(\sqrt{p} - r)$ . Using Pythagoras, we can form the equation

$(2r)^{2} = [2(p - r)]^{2} + [2(\sqrt{p} - r)]^{2} \Longrightarrow r^{2} = (p - r)^{2} + (\sqrt{p} - r)^{2}$ , (III).

Expanding this last equation and then solving the resulting quadratic for $r$ , we find that

$r = p + \sqrt{p} \pm \sqrt{2p\sqrt{p}}$ .

Assuming for now that $r \lt p$ , we have that $r = p + \sqrt{p} - \sqrt{2p\sqrt{p}}$ .

(Edit: If we were to choose the "+" sign then we would end up with a circle that has its center outside the first quadrant, and hence would render an invalid solution.)

After substituting this result into (I) and simplifying, we find that

$\sqrt{2p\sqrt{p}} - p = 2\sqrt{p}(\sqrt{2p\sqrt{p}} - \sqrt{p})$

$\Longrightarrow p - 2\sqrt{2}p\sqrt{\sqrt{p}} + \sqrt{2p\sqrt{p}} = 0$

$\Longrightarrow 2\sqrt{2}*y^{2} - y - \sqrt{2} = 0$ , where $y = \sqrt[4]{p}$ .

Noting that $y \gt 0$ we then find that $y = \dfrac{1 + \sqrt{17}}{4\sqrt{2}}$ , and so

$p = y^{4} = \dfrac{49 + 9\sqrt{17}}{128}$ .

Finally, $r = y^{4} + y^{2} - \sqrt{2}y^{3} = \dfrac{3(23 - \sqrt{17})}{128}$ ,

which is approximately $0.4424272$ , and so $\lfloor 10000*r \rfloor = \boxed{4424}$ .

soln