Circle in the Valley

A circle rests on the vertex of the curve \(y=x^2\). What radii can the circle have if it only touches the curve at one point? Here's a Desmos graph set up to play around with...

Solution

Force the circle \(x^2+(y-R)^2=R^2\) and the curve \(y=x^2\) to intersect at exactly one point.

Then replacing \(x^2\) with \(y\), we get \(y+(y-R)^2=R^2 \implies y+y^2-2Ry+R^2=R^2\\ \implies y^2+(1-2R)y=0\\ \implies y(y+1-2R)=0\).

\(y=0\) is a solution. The circle and curve are set up to intersect at this point.

The second bracket being zero gives \(y=-(1-2R)\) and we know \(y>0\) based on the initial setup so \(1-2R<0 \implies R>\frac{1}{2}\) for a solution.

 If \(y>0\) we do not want any further solutions, so we must have \(0<R\leq \frac{1}{2}\).

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Let's take this further. What about a circle resting in the valley of a cosine curve? Here's another Desmos graph with a circle on the curve \(y=1-cos(x)\) for convenience...

Solution

The previous method now involves solving equations containing both \(x\) and \(cos(x)\). Maybe reattempt the previous question with a different method prior to tackling this one.

Jumping back to the \(y=x^2\) question. We know the gradient of the circle and the curve are both 0 at the origin. Other than this point, provided the gradient of the circle is always greater than the gradient of the curve, the circle remains above the curve. Let's calculate the gradients.

The circle is \(f(x)=R-\sqrt{R^2-x^2}\) so \(f^{\prime}(x)= \frac{x}{\sqrt{R^2-x^2}}\) .
The curve has gradient \(2x\).

Force the gradient of the circle to be greater than the gradient of the curve.

\(\frac{x}{\sqrt{R^2-x^2}}<2x\) 

We know that \(x=0\) is where the gradients are equal, so ignore this point and say \(x\neq0\). Divide through by \(2x\) and multiply through by \(\sqrt{R^2-x^2}\) to get 

\(\frac{1}{2}>\sqrt{R^2-x^2} \implies R^2<\frac{1}{4}+x^2\)

This needs to hold true for \(0<x<R\). Very close to \(x=0\) gives \(R^2<\frac{1}{4} \implies R<\frac{1}{2}\).

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Jumping back to the \(y=1-cos(x)\) curve and applying this calculus method.

The circle is \(f(x)=R-\sqrt{R^2-x^2}\) so \(f^{\prime}(x)=\frac{x}{\sqrt{R^2-x^2}}\).
The curve is \(g(x)=1-cos(x)\) so \(g^{\prime}(x)=sin(x)\)

If \(f^{\prime}(x)>g^{\prime}(x)\) then \(\frac{x}{\sqrt{R^2-x^2}}>sin(x) \implies x>sin(x) \sqrt{R^2-x^2}\).

As \(x\) approaches \(R\), this is clearly true as \(\sqrt{R^2-x^2}\) tends to 0.
As \(x\) becomes very small, say \(\delta x\) then this becomes \(\delta x>sin(\delta x) \sqrt{R^2-{\delta x}^2}\) and using small angle approximations, we have \(\delta x>\delta x \sqrt{R^2-{\delta x}^2} \implies \sqrt{R^2-{\delta x}^2} <1 \implies R<1\).