Intersection points on circle, 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 and it intersects with the graph $f(x)=x^2$ , the above equation can be rewritten as $\displaystyle(x-h)^2+(x^2-k)^2=\frac{10}{\pi}$ . Expand it and we get $x^4-(2q-1)x^2-2px+\left(p^2+q^2-\frac{10}{\pi}\right)=0$

Compare the above equation with $\displaystyle g(x)-f(x)$ , we obtain

$\begin{aligned} 2q-1 &=& 3\\ -2p&=& b \\ p^2+q^2-\frac{10}{\pi} &=&1\\ \end{aligned}$

From the first equation, $\displaystyle q=2$ ; from the last equation, $\displaystyle p=\pm \sqrt{\frac{10}{\pi}-3}$ ; finally from the second equation, $\displaystyle b=2\sqrt{\frac{10}{\pi}-3}\approx 0.855801$ (as $b>0)$ . So $\displaystyle\lfloor{1000b}\rfloor = 855$ .