JEE Advanced 2021 — Paper 2 — Question 44

The largest circle in the family must touch the y-axis (x=0) and the parabola internally.

So let the circle centre be $(r,0)$ and radius $r$ (since it touches x=0).

Equation: $(x - r)^2 + y^2 = r^2$. The parabola is $y^2 = 4 - x$. Substitute $y^2$ from parabola into circle equation: $(x - r)^2 + (4 - x) = r^2$. Simplify: $x^2 - 2rx + r^2 + 4 - x = r^2 \rightarrow x^2 - (2r+1)x + 4 = 0$. For tangency, the quadratic in $x$ must have a single root, so discriminant $\Delta = 0$: $(2r+1)^2 - 16 = 0$. Solve: $2r+1 = 4$ (positive because $r>0$) \rightarrow $r = \frac{3}{2}$. Substitute $r$ into the quadratic: $x^2 - 4x + 4 = 0 \rightarrow (x-2)^2 = 0 \rightarrow x = 2$. Thus $\alpha = 2$.