<s>[INST] The radius of a circle, having minimum area, which touches the curve $y = 4 - x^2$ and the lines, $y = |x|$ is The radius of a circle, having minimum area, which touches the curve $y = 4 - x^2$ and the lines, $y = |x|$ is
My attempt is as follows:-
Let the circle be $x^2+y^2+2gx+2fy+c=0$, let the line $y=x$ touching circle at $(\alpha,\alpha)$
$$y\alpha+x\alpha+g(x+\alpha)+f(y+\alpha)+c=0$$
$$x(\alpha+g)+y(\alpha+f)+g\alpha+f\alpha+c=0$$
Comparing it with $y=x$
$$\dfrac{1}{\alpha+f}=\dfrac{-1}{\alpha+g}$$
$$\alpha+g=-\alpha-f=2\alpha+g+f=0\tag{1}$$
$$g\alpha+f\alpha+c=0$$
$$g+f=-\dfrac{c}{\alpha}$$
$$-2\alpha=-\dfrac{c}{\alpha}$$
$$2\alpha^2=c\tag{2}$$
Comparing it with $y=-x$, let the line $y=-x$ touching circle at $\left(\beta,-\beta\right)$
$$-y\beta+x\beta+g(x+\beta)+f(y-\beta)+c=0$$
$$x(g+\beta)+y(f-\beta)+g\beta-f\beta+c=0$$
$$\dfrac{1}{f-\beta}=\dfrac{1}{g+\beta}$$
$$f-g=2\beta$$
$$g\beta-f\beta+c=0$$
$$\beta(g-f)=-c$$
$$2\beta^2=c$$
So we got $\alpha=\pm\beta$
Let the parametric point on $y=4-x^2$ be $(\gamma,4-\gamma^2)$
$$y-(4-\gamma^2)=-2\gamma(x-\gamma)$$
$$y-4+\gamma^2=-2\gamma x+2\gamma^2$$
$$y-4=-2\gamma x+\gamma^2$$
Let's write the equation of tangent to circle at point $(\gamma,4-\gamma^2)$
$$y(4-\gamma^2)+x\gamma+g(x+\gamma)+f(y+4-\gamma^2)+c=0$$
$$y(4-\gamma^2+f)+x(\gamma+g)+g\gamma+f(4-\gamma^2)+c=0$$
$$\dfrac{1}{4-\gamma^2+f}=\dfrac{2\gamma}{\gamma+g}$$
Its getting too long, any other way of doing this question?
 [/INST] 
Given the symmetry with respect to the $y$-axis,let the equation of the circle be
$$x^2+(y-b)^2=r^2$$
Note that the origin, the center of the circle and the touch point with one of the lines $y=|x|$ form a 45-45 right triangle, which gives 
$$ b = \sqrt2 r\tag 1$$
Also, match the gradients at the touch point with the curve $y=4-x^2$,
$$ y' = -2x = -\frac{x}{y-b}$$
which leads to $y = \frac12 + b$. Plug it into $y=4-x^2$ to get $x$-coordinates of the touch points $x=\pm\sqrt{\frac72-b}$. Then, the distance of the center to the touch points is
$$(\frac72 -b) + \frac14 = r^2=\frac{b^2}2$$
where (1) is used in the last step. solve to obtain $b  = \sqrt{\frac{17}2}-1$ 
and, in turn, the radius 
$$r = \frac12(\sqrt{17}-\sqrt2)$$
</s>