Orthogonal parabolas, part 2

Suppose the equation of the circle is $\displaystyle (x-h)^2+(y-k)^2=r^2$ .

Since the circle's area is 10, the above equation can be rewritten as $x^2+y^2-2hx-2ky+h^2+k^2 -\frac{10}{\pi}=0$

From the two parabolas, $\displaystyle\begin{cases} y=2x^2+a \\ x=2y^2+b \end{cases}$ , the difference gives $x^2+y^2-\frac{1}{2}x-\frac{1}{2}y+\frac{a+b}{2}=0$

Compare these two equations, we obtain $\displaystyle h=k=\frac{1}{4}$ and $\displaystyle a+b=2\left(h^2+k^2-\frac{10}{\pi}\right)=\frac{1}{4}-\frac{20}{\pi}$ .

Since $a,b<0$ , $\displaystyle |a|+|b| =-(a+b)=\frac{20}{\pi}-\frac{1}{4}$ . From AM-GM, we have $\displaystyle ab=|ab|\le \left(\frac{|a|+|b|}{2}\right)^2=\left(\frac{10}{\pi}-\frac{1}{8}\right)^2 \approx 9.35197$ .

So the maximum value of $\displaystyle\lfloor{1000ab}\rfloor=9351$ .