Circle buddies in a rectangle

Problem requiring patience :)

Let the radius of each circle be $r$ . Then the equation of $E_1$ , considering the origin of coordinates at $C$ , is

$(x-r) ^2+(y-r) ^2=r^2$

Equation of $E_2$ is

$(x-2r)^2+(y-r) ^2=r^2$

Solving we get the coordinates of $I$ as

$x_I=\dfrac {3r}{2},y_I=\dfrac {(2+\sqrt 3)r}{2}$

Equation of red circle is

$(x-\frac{3r}{2})^2+(y-\frac{(2+\sqrt 3)r}{2})^2=r^2$

Solving this equation with the equation of $E_1$ we get the coordinates of $M$ as

$x_M=\dfrac r2,y_M=\dfrac {(2+\sqrt 3)r}{2}$

Since $|\overline {JK}|$ is minimum, this line segment must be perpendicular to $\overline {AB}$

So, coordinates of $J$ are

$x_J=\dfrac {3r}{2},y_J=2r$ , and of $K$ are

$x_K=\dfrac {3r}{2},y_K=\dfrac {(4+\sqrt 3)r}{2}$

So, area of $\triangle {JMK}$ is

$\dfrac 12|\frac r2(\frac{4+\sqrt 3}{2}r-2r)+\frac{3r}{2}(2r-\frac{2+\sqrt 3}{2}r)+\frac{3r}{2}(\frac{2+\sqrt 3}{2}r-\frac{4+\sqrt 3}{2}r)|$

$=\dfrac {r^2\sqrt 3}{4}$

It is given that $r=\dfrac {10}{2}=5$

So, area of the triangle is $\dfrac {25\sqrt 3}{4}$

Since the triangles $\triangle {JMK}$ and $\triangle {JLK}$ have equal area, therefore area of the quadrilateral $LJMK$ is twice the sum of $\triangle {JMK}$ :

Area $=\dfrac {25\sqrt 3}{2}$

Hence $X=25,Y=3,Z=2$

The equation of the given circle is

$(x-2)^2+(y-3)^2=25=5^2$

So the coordinates of the uppermost point of this circle are

$x_u=2,y_u=3+5=8$ , and their product is $x_uy_u=2\times 8=\boxed {16}$ .