Dynamic Geometry: P97

Let an arbitrary point on the locus be $P(x,y)$ and the distance of the center either circle to the origin $(0,0)$ of the $xy$ -plane be $a$ . By Pythagorean theorem ,

$(x+a)^2 + y^2 = 1^2 \implies y = \sqrt{1-(x+a)^2}$

And the area of the rectangle is $A=4xy=4x\sqrt{1-(x+a)^2}$ . To find the relation between $x$ and $y$ , when $A$ is maximum,

$\begin{aligned} \frac {dA}{dx} & = 4 \sqrt{1-(x+a)^2} - \frac {4x(x+a)}{\sqrt{1-(x+a)^2}} & \small \blue{\text{Putting }\frac {dA}{dx}=0} \\ \sqrt{1-(x+a)^2} & = \frac {x(x+a)}{\sqrt{1-(x+a)^2}} \\ 1 - (x+a)^2 & = x(x+a) & \small \blue{\text{or } y^2 = x(x+a)} \\ (x+a)^2 + x(x+a) - 1 & = 0 \\ a^2 + 3ax + 2x^2 - 1 & = 0 & \small \blue{\text{Solving the quadratic for }a} \\ \implies a & = \frac {\sqrt{x^2+4}-3x}2 & \small \blue{\text{Since }a > 0} \end{aligned}$

Since $y^2 = x(x+a) = x^2 + x \cdot \dfrac {\sqrt{x^2+4}-3x}2 = \dfrac {\sqrt{x^2+4}-x^2}2$ ,

$\implies y = \sqrt{\frac {\sqrt{x^2+4}-x^2}2}$

Therefore $p+q = 4+2 = \boxed 6$ .