Find the radius of the largest circle that lies above the x-axis and below the parabola $y = 2 − x ^2$ .

Question

Find the radius of the largest circle that lies above the x-axis and below the parabola $y = 2 − x ^2$

I do not know how to approach this problem, I guess we already have (0,0) that intersects the circle and 2 other points that are tangent to 2 lines that have the vertex of the parabola at (0,2)? I'm not sure how to get the points.

Answer 1

You are actually looking for the inscribed circle of a spade-shape region. This can be accomplished without resort to derivatives (but you need to know how to minimize a quadratic function).

Your first observation is correct since by symmetry the lowest point on the circle must be on the origin in order to maximize the radius. Then we can set the circle to be centered at $C=(0,r)$ with radius $r$. Obviously, the tangent lines passing two other intersection points are simultaneously tangent lines of the circle and the parabola. Notice that if this is true, $r$ must be the shortest distance from $C$ to the parabola. Thus we can find the intersection points $(\pm x_0,y_0)$ by minimizing $r^2=x^2+(y-r)^2$ with $y=2-x^2$. Then $r$ can be solved out by plugging back $(\pm x_0,y_0)$ into $r^2=x^2+(y-r)^2$.

I will leave the explicit calculation for you. : )

Answer 2

For blue circle, coordinates of point $B$ are $B(x_B,y_B)$ with:

$$x_B=r\sin\alpha$$

$$y_B=r+r\cos\alpha$$

The point B lies on the parabola:

$$y_B=2-x_B^2$$

$$r+r\cos\alpha=2-r^2\sin^2\alpha\tag{1}$$

At the same time $AB$ is perpendicular to common tangent at point $B$ which translates to:

$$\alpha+\beta=\pi$$

$$\tan\alpha=-\tan\beta=-y_B'=2x_B$$

$$\frac{\sin\alpha}{\cos\alpha}=2r\sin\alpha$$

At this point we will assume that $\sin\alpha\ne0$. The case $\sin\alpha=0$ is actually the case of the red circle and can be easily handled later.

$$r\cos\alpha=\frac12\tag{2}$$

From (2):

$$\cos\alpha=\frac1{2r}$$

$$\sin^2\alpha=1-\cos^2\alpha=1-\frac1{4r^2}$$

Replace this into (1) and you get:

$$r+r\frac1{2r}=2-r^2(1-\frac1{4r^2})$$

$$r+\frac12=2-r^2+\frac14$$

$$r^2+r-\frac74=0$$

Solve for $r$ and take the positive solution only:

$$r=\frac{2\sqrt2-1}{2}\approx0.914$$

...which is well in agreement with the measurment obtained from Geogebra. Increase the radius just a little bit and the circle will intersect the parabola. Increase it even more and you get the red circle (case $\sin\alpha=0$) which is also not a solution.

For some more "obtuse" parabola, the process from the above could lead to two complex solutions for $r$. That would leave us with $\sin\alpha=0$ (the red circle) as the only solution. But this is not the case here.