The radius of a circle, having maximum area, which touches the curve $y = 4 – x^2$ and the lines, $y = |x|$ is

Question

The radius of a circle, having maximum area, which touches the curve $y = 4 – x^2$ and the lines, $y = |x|$ is

Note this is different from: this which is asking for minimum area.

If you click on the link and see the comments section of the answer the person said the area of the circle with maximum area touches the parabola at (0,4) and intersects the parabola at two other points and along with this touches the curve y = |x| at two distinct points.

My question is: isn't the circle with maximum area the one touching the parabola at (0,4) and touching the curve y=|x| at (0,0)? Can we call such a circle touching both y=|x| and the parabola?

Here is a diagram:

Here purple is the line y =|x|, green is the parabola, black is the circle with minimum radius (which I agree with) and red is the one they are claiming to be circle with the maximum radius which I'm not sure. Shouldn't it be the circle $x^2 + (y-2)^2 = 4$? That circle touches y=|x| at (0,0) and parabola at (0,4). The only difference is the derivative at the touching point isn't the same. What is right? Please clarify.

Answer 1

The answer should be the circle touching y=|x| and y=4-$x^2$ from above with r= 4(1+$\sqrt2$). Its radius is larger than 4($\sqrt2$-1), which is the radius of the other circle touching the parabola from underneath.

As shown in the graph, let r= radius of the circle centered at o'(0,x+r) , the circle touches y=x at A(x,x).

With $\Delta$ OO'A, $(4+r)^2$=$r^2$+$(\sqrt2x)^2$

With $\Delta$ BO'A, $(4+r-x)^2+x^2$=$r^2$

Solving these two equations,

x($x^2-8x+8)=0, $r=$\frac{x^2-8}{4}$

Only solution x=4+2$\sqrt2$, r=4(1+$\sqrt2$) provides positive r.

Answer 2

The equation of the circle touching $y=\pm x$ and center on $y$-axis be $$x^2+(y-h)^2=h^2/2~~~(1)$$ Now if ith has to touch $y=4-x^2$, then the equation $$4-y+(y-h)^2=h^2/2 \implies y^2-(2h+1)y+h^2/2+4=0$$ should have a unique root of $y$. Imposing $B^2=4AC$, we get $$2h^2+4h-15=0 \implies h=\frac{-4\pm \sqrt{136}}{4}$$ Choosing positive value we have $h=\frac{\sqrt{136}-4}{4}$ in (1).

For the other possibility, (1) should give $y(0)=4$ and $h<4$. This gives $(4-h)^2-h^2/2 \implies h(1\pm 1/\sqrt{2})=4$, we get $$h=\frac{4 \sqrt{2}}{\sqrt{2}\pm1}~~~~(2)$$, in (1), choosing $+$ sign.

EDIT So maximum radius from (2) $$r=h/\sqrt{2}=\frac{4}{\sqrt{2}-1}.$$ as found by @Bright Star, below.

Answer 3

Two curves $y=f(x)$ and $y= g(x)$ are said to be touching at $(x_0,y_0)$ if $y_0=f(x_0)=g(x_0)$ and $f'(x_0)=g'(x_0)$.

In your case: Indeed the circle $x^2 + (y-2)^2 = 4$ and straight lines $y=|x|$ meet at $(0,0)$ but the function $y=|x|$ does not have a unique tangent at $(0,0)$.