The radius of the circle having minimum area which touches the curve y=4-x² and the lines, y=|x| is?
I tried using the normal to the curve and satisfying it with the centre of the circle but it just gives me a complex 4th degree equation which I'm unable to solve...
Do you see why you need to solve $|x|=4-x^2$?
The symmetry of the problem suggests the center of the circle lies on the y axis.
Minimization implies the circle is tangent to some of the boundaries in question.
That the circle has to be tangent to $y=|x|$ places some restrictions on the radius.
Solving the first equation gives you a range to look into.
There is not enough constraints to choose a unique circle. Here is a desmos graph of the set up:
To make this question work and get the intended answer as given by the JEE authorities, we need to say that the circle touches both $y=x$ and $y=-x $ of $|x|$ function.
As mentioned in above answer, by considering the fact that the circle(radius=r) will be tangent to $y=x$ and $y=-x$ the centre of the circle can be found out to be $(0,\sqrt{2}r)$ [See figure]
Therefore the equation of the circle will be $$x^2+(y-\sqrt{2}r)^2=r^2$$
the variable r can be eliminated by using the condition that it is tangent to $y=4-x^2$
