Falzoon's Ellipse

First, let the radius of the big circle be $1$ . Then let

$r$ be the radius of the small circle
$R$ be the major axis of the ellipse
$k$ be the shrink factor of the ellipse

We set up the equations of the circles and ellipse such that the bottom of the ellipse is at the origin $(0,0)$ tangent to the small circle which is itself tangent to the large circle at $(0,-2r)$ . Then to find the intersection of the ellipse and the big circle, we solve the equation

$(1-2r)+\sqrt { 1-{ x }^{ 2 } } =k(R+\sqrt { { R }^{ 2 }-{ x }^{ 2 } } )$

Let $B=(1-2r)-kR$ to reduce clutter

Expanding and re-arranging, we have the quadratic in ${ x }^{ 2 }$ to solve

$(1-2{ k }^{ 2 }+{ k }^{ 4 }){ x }^{ 4 }+(-2+2{ B }^{ 2 }+2{ k }^{ 2 }+2{ B }^{ 2 }{ k }^{ 2 }+2{ R }^{ 2 }{ k }^{ 2 }-2{ R }^{ 2 }{ k }^{ 4 }){ x }^{ 2 }+(1-2{ B }^{ 2 }+{ B }^{ 4 }-2{ k }^{ 2 }{ R }^{ 2 }-2{ B }^{ 2 }{ k }^{ 2 }{ R }^{ 2 }+{ k }^{ 4 }{ R }^{ 4 })=0$

For the intersections to be points of tangency, the radical in the quadratic solution must vanish, which means this condition must be satisfied

${ (-2+2{ B }^{ 2 }+2{ k }^{ 2 }+2{ B }^{ 2 }{ k }^{ 2 }+2{ R }^{ 2 }{ k }^{ 2 }-2{ R }^{ 2 }{ k }^{ 4 }) }^{ 2 }-4(1-2{ k }^{ 2 }+{ k }^{ 4 })(1-2{ B }^{ 2 }+{ B }^{ 4 }-2{ k }^{ 2 }{ R }^{ 2 }-2{ B }^{ 2 }{ k }^{ 2 }{ R }^{ 2 }+{ k }^{ 4 }{ R }^{ 4 })=0$

Expanding and re-arranging, this reduces to

$16{ B }^{ 2 }{ k }^{ 2 }(-1+{ B }^{ 2 }+k^{ 2 }+{ R }^{ 2 }-{ k }^{ 2 }{ R }^{ 2 })=0$

Plugging $B=(1-2r)-kR$ back in, we end up with the quadratic to solve for $k$

${ k }^{ 2 }-2(1-2r)Rk+({ R }^{ 2 }-4r(1-r))=0$

which gets us

$k=(1-2r)R+2\sqrt { r(1-r)(1-{ R }^{ 2 }) } )$

The area of the ellipse is pi times $k{ R }^{ 2 }$ . We differentiate this with respect to $R$ to find the maximum. After re-arranging, we have the quadratic in ${ R }^{ 2 }$ to solve

$9{ R }^{ 4 }+3(3-2r)(2r+1){ R }^{ 2 }+16r(1-r)=0$

which gets us

$R=\frac { 1 }{ \sqrt { 6 } } \sqrt { (3-2r)(2r+1)-(2r-1)\sqrt { 4{ r }^{ 2 }-4r+9 } }$

Then we solve by numerical means the equation

$k{ R }^{ 2 }-{ r }^{ 2 }=0$

to find that $r=0.5634366252335449…$

From this, the answer to this problem is

Floor $\left\lfloor 10000(2{ r }^{ 2 }) \right\rfloor =6349$

Equating x^2 and d y/ d x of big circle and ellipse makes the answer.

Fix small circle at x^2 + (y - 1)^2 = 1 where radius of 1 unit is taken as reference for ratios. Take x^2/ a^2 + (y - b - 2)^2/ b^2 = 1 for ellipse. Since Pi a b = Pi r^2 and r = 1, 1/ a = b.

x^2 + (y - b - 2)^2/ b^4 = 1/ b^2 is the ellipse above the circle x^2 + (y - 1)^2 = 1 .

x^2 + (y - h)^2 = h^2 is taken for big circle.

Subtract big circle by ellipse:

(b^4 - 1) y^2 - 2 (b^4 h - b - 2) y - 4 (b + 1) = 0

Differentiate big circle x^2 + y^2 - 2 h y = 0,

d y/ d x = x/ (h - y)

Differentiate ellipse b^ 4 x^2 + (y - b - 2)^2 = b^2,

dy/ d x = b^4 x/ (2 + b - y)

Equate tangent of ellipse to tangent of big circle,

y = (b^4 h - b - 2)/ (b^4 - 1)

Substitute y = (b^4 h - b - 2)/ (b^4 - 1) into (b^4 - 1) y^2 - 2 (b^4 h - b - 2) y - 4 (b + 1) = 0,

(b^4 h - b - 2)^2/ (1 - b^4) = 4 (1 + b)

=> h = (2 + b +/- 2 sqrt((1 - b^4)(1 + b)))/ b^4

d h/ d b = 0 for minimum h

=> b +/- (b - 4 b^4 - 5 b^5)/ sqrt((1 - b^4)(1 + b)) = 8 + 4 b +/- 8

sqrt((1 - b^4)(1 + b))

Since removing surd form will make polynomial of high degree which must only be solved by iteration method for real roots, knowing 0 < b < 1 and 1 + b < h < 2, direct substitution can solve this equation. Found that only b - (b - 4 b^4 - 5 b^5)/ sqrt((1 - b^4)(1 + b)) = 8 + 4 b - 8 sqrt((1 - b^4)(1 + b)) is changing sign between b = 0.7 and b = 0.8 for the range when arranged,

b = 0.724911899756926

hence h = 1.77482250037526

Maximum (ellipse + small circle)/ (big circle) = [Pi (a b + r^2)]/ [Pi h^2] = 2/ h^2 = 0.634921661309133

Floor(0.634921661309133 x 10000) = 6349