The area of the blue circle centered at the origin, which is inscribed in the parabola $y = x ^2 − 100$, can be expressed as $\frac{a}{b}\pi$

Question

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 $\frac{a}{b}\pi$.

∴$a= 399 b= 4$

∴$(a+b)=399+4=403$

Answer 1

Your answer is correct, but there's a couple of small mistakes in your working.

When you set the derivative equal to $0$ you concluded that either $2x^3 - 199x = 0$ or $\sqrt{x^4 - 199x + 10000} = 0$. The second of these equations is not correct. Remember: a fraction equals $0$ when its numerator is zero and its denominator is non-zero. After setting the derivative equal to $0$, the conclusion you should get is $$2x^3 - 199x = 0\ \text{and}\ \sqrt{x^4 - 199x + 10000} \neq 0$$

You go on to make the correct conclusion that the derivative equals $0$ at $x = 0$ and $x = \pm \sqrt\frac{199}{2}$. However, you need to justify why you then disregard $x = 0$. You could at least mention that it's obvious from the diagram that the quadratic doesn't touch the circle at $x=0$ (although diagrams can't always be trusted). A check of the second derivative would confirm that this is a local maximum of the distance from the origin, and we're looking for minima.

Answer 2

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 $4r^2 = 399$.