JEE Advanced 2021 Paper 2 · Q07 · Parabola

Correct answer:1.5

Solution

Let circle $C$: $(x-r)^2 + y^2 = r^2$. Substituting $y^2 = 4-x$ gives $(x-r)^2 + 4 - x = r^2$, i.e. $x^2 - (2r+1)x + 4 = 0$. Tangency requires discriminant zero: $(2r+1)^2 = 16$, so $2r+1 = 4$, $r = 1.5$.

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