<s>[INST] The area of the blue circle centered at the origin, which is inscribed in the parabola $y = x ^2 − 100$, can be expressed as $\frac{a}{b}\pi$ The area of the blue circle centred at the origin, which is inscribed in the parabola $$y = x^{2}-100$$ can be expressed as $\frac{a}{b}\pi$, where $\frac{a}{b}$
is a
fraction in its lowest terms (i.e. a, b are coprime positive integers). What is
the value of  $a + b$?

I did this far. Can anyone please check and let me know if its correct or not?
Here the equation of Parabola is,$$y=x^2−100$$
As, the blue circle is centered at the origin so the circle equation would be,$$x^{2}+y^{2}=r^{2}$$
As, the circle is inscribed in the parabola the radius of the circle is equal to the shortest distancefrom the origin to the any point on the parabola
∴The radius of the circle,$$r=\sqrt{x^{2}+y^{2}}=\sqrt{x^{2}+(x^{2}-100)^2}$$
By differentiating r with respect to x we get,
$$\frac{dr}{dx}=\frac{d}{dx}\sqrt{x^2+(x^2-100)^2}\\\Rightarrow\hspace{.1cm}\frac{dr}{dx}=\frac{2x^3-199x}{\sqrt{x^4-199x^2+10000}}$$
Let,$$\frac{dr}{dx}= 0$$
So,$\hspace{.1cm}2x^3−199x= 0$   or  $\sqrt{x^4-199x+10000}= 0$
$⇒x=  0,±\sqrt{\frac{199}{2}}$
$\sqrt{x^4-199x+10000}$  has no real solutions for x.
∴$x=  0,±\sqrt{\frac{199}{2}}$
∴If x=±\sqrt{\frac{199}{2}},
∴$y^2=x^2−100=  (±\sqrt{\frac{199}{2}})^2−100\hspace{.5cm}\text{[Implementing x value]}\\=−12$
∴At $x=\pm\sqrt{\frac{199}{2}}$,
radius, $r=\sqrt{x^2+y^2}\\⇒r=\frac{\sqrt{399}}{2}$
∴The area of the circle is= $\displaystyle πr^2=π(\frac{\sqrt{399}}{2})^2=\frac{399\pi}{4}$
The circle area was expressed as $\frac{a}{b}\pi$.
∴$a= 399 b= 4$
∴$(a+b)=399+4=403$
 [/INST] You can combine your first two equations and get
$     y^2 + y + 100 - r^2 = 0$. Then
$$y = \frac{-1 \pm \sqrt{1 - 400+4r^2} }{2}$$If the circle intersects the parabola at two points - single value of y - then the square root term must be $0$, or $4r^2 = 399$.
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