Tangencies

Let the radius of the circle be $r$ . Then the equation of the circle is $(x-3+r)^2 + (y-2+r)^2 = r^2$ . Consider the point of contact $P(u,v)$ between the circle and the ellipse. Other than satisfying the equations of the ellipse and circle, the gradients of both the ellipse and circle are equal. Therefore

$\begin{cases} 2 \cdot \dfrac x9 + 2 \cdot \dfrac y4 \cdot \dfrac {dy}{dx} = 0 & \implies \dfrac {dy}{dx} = - \dfrac {4x}{9y} \\ 2(x-3+r)+2(y-2+r) \dfrac {dy}{dx} = 0 & \implies \dfrac {dy}{dx} = - \dfrac {x-3+r}{y-2+r} \end{cases} \implies \dfrac {4x}{9y} = \dfrac {x-3+r}{y-2+r}$

Now $P(u,v)$ satisfies the following system of equations

$\begin{cases} \begin{aligned} \frac {x^2}9 + \frac {y^2}4 & = 1 & ...(1) \\ (x-3+r)^2 + (y-2+r)^2 & = r^2 & ...(2) \\ \frac {x-3+r}{y-2+r} & = \frac {4x}{9y} & ...(3) \end{aligned} \end{cases}$

Since there are three equations and three unknowns, we can solve the system of equations. What I did was, using equation $(1),$ first expressed $y$ in term of $x$ as $y = \sqrt{4(1-\frac {x^2}9)}$ . Using the corresponding values of $x$ and $y$ calculated the value of $r$ using equation $(3)$ as $r = \frac {5xy+8x-27y}{4x-9y}$ . Then went to find the corresponding values of $x$ , $y$ , and $r$ that satisfied equation $(2).$ I did this with an Excel spreadsheet with manual iterations. Perhaps @Pi Han Goh can come up with an algebraic solution. I have neither the skills nor the patience.

As if struck by good luck, else I would have to fail @Fletcher Mattox this time, it was found that $u^2 = 5.6$ and $v^2 = 1.6$ or $u=3\sqrt{\frac 35}$ and $v = 2\sqrt{\frac 25}$ . Substituting the values into

$r = \frac {5uv+8u-27v}{4u-9v} = 5 + \sqrt 6 - \sqrt {10} - \sqrt{15}$

Therefore $a+b+c+d = 5+6+10+15 = \boxed{36}$ .

If $r$ is the radius of the corner circle, then its center is

$C = (3 - r, 2 - r)$

Let $t$ be the eccentric angle of the ellipse, so that points on the ellipse can be parameterized as

$P = ( 3 \cos t, 2 \sin t )$

Two conditions must hold for the point of tangency $P$ :

1. The normal vector at $P$ has the same slope as $PC$ .
2. $PC$ has a length of $r$ .

Since the normal vector at $P$ is along $(2 \cos t, 3 \sin t)$ , the first requirement translates into,

$\dfrac{ 3 \sin t }{ 2 \cos t } = \dfrac{ (2 - r) - 2 \sin t}{ (3 - r) - 3 \cos t}$

which, after re-arranging, becomes,

$3 \sin t ( (3 - r) - 3 \cos t ) - 2 \cos t ( (2 - r) -2 \sin t ) = 0$

so that,

$\mathbf{f}_1 = - 2 (2 - r) \cos t + 3 (3 - r) \sin t - 5 \sin t \cos t = 0$

And the second requirement, becomes,

$\mathbf{f}_2 = ((3 - r) - 3 \cos t )^2 + ((2 - r) - 2 \sin t )^2 - r^2 = 0$

This is a nonlinear system of two equations in two unknowns, $\mathbf{f(x) } = \mathbf{0}$

It can be readily solved using Newton-Raphson's method (multivariate). My implementation converged in 3 iterations

to $r = 0.414228736407382$ and $t = 0.684719203002281$

Next thing, using a trial-and-error loop, I found that $r = 5 + \sqrt{6} - \sqrt{10} - \sqrt{15}$

Hence, the answer is $5 + 6 + 10 + 15 = \boxed{36}$

I've checked other cases, and it seems that for the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ , the radius of the circle is given by $a + b + \sqrt{ab} - \sqrt{a(a + b)} - \sqrt{b(a + b)}$ .