Maximum area of a circle bound by a line and a parabola

Question

I had a math exam today and I couldn't solve this problem. How are you supposed to calculate the maximum area of the circle c with only algebra?

Answer 1

Consider the following inequality: \begin{align*} \left(x-\frac{1}{2}\right)^2\geq0\\ x^2-x+\frac{1}{4}\geq 0\\ x^2\geq x-\frac{1}{4} \end{align*} Note that equality is only reached when $x=\frac{1}{2}$. Geometrically, we can interpret this inequality as follows: the line $y=x-\frac{1}{4}$ is tangent to the graph $y=x^2$ at $x=\frac{1}{2}$.

After drawing a graph like the one above, it becomes apparent that the diameter of circle $C$ is at most the perpendicular distance between the two lines $y=x$ and $y=x-\frac{1}{4}$. This distance is equal to $\frac{\sqrt{2}}{8}$, making the maximum area of the circle $\frac{\pi}{128}$.

EDIT: It may be possible that the circle I mentioned before in my answer does not lie entirely above the parabola. To check this, we need to make sure that the radius of the osculating circle of the parabola at $x=\frac{1}{2}$ is greater than or equal to $\frac{\sqrt{2}}{16}$. We can calculate the radius of the osculating circle using the following formula: $$R=\frac{\left(1+y'^2\right)^{3/2}}{y''}=\frac{\left(1+2x\right)^{3/2}}{2}=\frac{2^{3/2}}{2}=\sqrt{2}>\frac{\sqrt{2}}{16}$$ This is actually not sufficient enough either, since the osculating circle at $x=\frac{1}{2}$ actually goes below the parabola: