JEE Advanced 2021 — Paper 2 — Question 43

The region is $R = \{(x,y): x \ge 0,\; y^2 \le 4 - x\}$.

The largest circle in $R$ with centre on the x-axis must touch the parabola $y^2 = 4 - x$ tangentially.

Let its centre be $(r,0)$ and radius $r$, so its equation is $(x - r)^2 + y^2 = r^2$.

Substituting $y^2 = 4 - x$ gives $(x - r)^2 + 4 - x = r^2$.

Simplify: $x^2 - 2rx + r^2 + 4 - x = r^2$ → $x^2 - (2r + 1)x + 4 = 0$.

For tangency, the discriminant must be zero: $(2r + 1)^2 - 16 = 0$.

Hence $2r + 1 = 4$ (positive root) → $r = \frac{3}{2} = 1.5$.

The radius of circle C is $1.5$.