Dynamic Geometry: P33

See the following diagram: - The moving red point is point $A\left(a;0\right)$ on our diagram. We set the radius of the circle as $1$

• We use the pythagorean theorem to express the radius $r$ of the cyan circle in terms of $a$

• We need to solve the equation: $2r^2+a^2=\left(1-r\right)^2$ $<=>$ $r=\sqrt{2-a^2}-1$

• We know that the center of the cyan circle has coordinates $\left(a;r\sqrt{2}\right)$ which after substitution becomes $\left(a;\sqrt{4-2a^2}-\sqrt{2}\right)\:$

• Having the $x$ and $y$ coordinate of the center in terms of $a$ , we use substitution to get $y$ in terms of $x$ to get the equation of the locus.

• We get: $y=\sqrt{4-2x^2}-\sqrt{2}\:$

• Obviously the area of the cyan circle is its maximum when point A has coordinates $\left(0;0\right)$ . The area is then: $\pi \left(3-2\sqrt{2}\right)$

• Now we evaluate the area bounded by the two pink curves: $2\cdot \int _{-1}^1\sqrt{4-2x^2}-\sqrt{2}=\dfrac{2\pi -4}{\sqrt{2}}$

• Then we evaluate the ratio between those areas and get: $\dfrac{\pi \left(3\sqrt{2}-4\right)}{2\pi -4}$

• $d=3$ , $b=2$ , $c=4$ and $\left|\sqrt{3+2+4}\right|=\left|\sqrt{9}\right|=3$

Let the radius of the big circle (I don't see much black but white) be $1$ and its center be $O(0,0)$ of the $xy$ -plane with the slanting green line (Does a circle has a diagonal?) as the $x$ -axis. Turn you head $45^\circ$ and I will show you not only we can prove that the pink locus is part of an ellipse but how.

Let an arbitrary point on the locus be $P(x,y)$ and the radius of the circle be $r$ when $P$ is the center of the variable circle (yah, I know it is cyan, but to be politically-correct to the color-vision impaired). By Pythagorean theorem we have:

$\begin{cases} x^2 + y^2 = OP^2 = (1-r)^2 & ...(1) \\ y^2 = r^2 + r^2 = 2r^2 & ...(2) \end{cases}$

$\begin{aligned} (1): \quad x^2 + y^2 & = r^2 - 2r + 1 & \small \blue{\text{From }(2): \ \ r = \frac y{\sqrt 2}} \\ x^2 + y^2 & = \frac {y^2}2 - \sqrt 2 y + 1 \\ x^2 + \frac {y^2}2 + \sqrt 2 y & = 1 \\ x^2 + \left(\frac y{\sqrt 2} + 1 \right)^2 & = 2 \\ \frac {x^2}2 + \frac {(y+\sqrt 2)^2}4 & = 1 \end{aligned}$

See! The locus is the top part of an ellipse with center at $(0, - \sqrt 2)$ , minor-axis $a=\sqrt 2$ and major-axis $b=2$ (Note that $a$ and $b$ has nothing to do with the answer option $\sqrt{a+b+c}$ . Of course this work out to be $y = \sqrt{4-2x^2} - \sqrt 2$ as shown by @Valentin Duringer , and we can find the area bounded the two curves by using $2 \int_{-1}^1 y\ dx$ . But since it is an ellipse, we can apply:

$\begin{cases} \dfrac {x^2}2 = \cos^2 \theta & \implies x = \sqrt 2 \cos \theta \\ \dfrac {(y+\sqrt 2)^2}4 = \sin^2 \theta & \implies y = 2 \sin \theta - \sqrt 2 \end{cases}$

And if you don't want to find the definite integral using Wolfram Alpha or other software, you can do as below:

$\begin{aligned} A & = 2 \int_{-1}^1 y\ dx = 4 \int_0^1 y\ dx & \small \blue{\text{Since }y \text{ is even.}} \\ & = 4 \int_\frac \pi 2^\frac \pi 4 (2 \sin \theta - \sqrt 2) \blue{(-\sqrt 2 \sin \theta)\ d \theta} & \small \blue{\text{As }dx = -\sqrt 2 \sin \theta} \\ & = \int_\frac \pi 4^\frac \pi 2 \left(8\sqrt 2 \sin^2 \theta - 8\sin \theta \right) d\theta \\ & = \int_\frac \pi 4^\frac \pi 2 \left(4\sqrt 2 (1 - \cos 2 \theta) - 8\sin \theta \right) d\theta \\ & = 4\sqrt 2 \theta - 2\sqrt 2 \sin 2 \theta + 8 \cos \theta \ \bigg|_\frac \pi 4^\frac \pi 2 \\ & = \sqrt 2 \pi + 2 \sqrt 2 - 4 \sqrt 2 \\ & = \sqrt 2 \pi - 2 \sqrt 2 \end{aligned}$

The largest variable circle is when $x=0$ , $\implies \theta = \frac \pi 2$ , $y = 2 - \sqrt 2$ , and $r = \sqrt 2 -1$ . Then the area of the largest variable circle $A_\bigcirc = \pi (\sqrt 2 -1)^2 = (3-2\sqrt 2)\pi$ . And the ratio:

$\frac {A_\bigcirc}A = \frac {(3-2\sqrt 2)\pi}{\sqrt 2 \pi - 2\sqrt 2} = \frac {(3\sqrt 2 -4)\pi}{2\pi - 4}$

Then the require answer (I can't mention $a$ and $b$ because I have mentioned them for other purposes) is $\sqrt{3+2+4} = \boxed 3$ . (Sorry for my sarcasm, I'm just trying to be a more serious amateur math nerd.)