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Concentric circles where tangents from C2 to C1 meet C2 at ends of a diameter - JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
Let C1C_1 and C2C_2 be two concentric circles of radii r1r_1 and r2r_2 (r1<r2)(r_1<r_2). Suppose that from every point of C2C_2 the two tangents drawn to C1C_1 meet C2C_2 again at the ends of a diameter of C2C_2. Then
Ar2=2r1r_2 = 2r_1
Br2=2 r1r_2 = \sqrt2\, r_1correct
Cr22<2r12r_2^2 < 2r_1^2
DC2C_2 is director circle of C1C_1correct
Solution
Question attachment Step 1: AA, PP, BB lie on C2C_2 and ABAB is a diameter of C2C_2, so by the angle in a semicircle,
∠APB=90∘\angle APB = 90^\circ
 Step 2: PAPA and PBPB are the two tangents from PP to C1C_1, so this pair of tangents is perpendicular. The locus of points from which the two tangents to a circle are perpendicular is the director circle of that circle, and every point PP of C2C_2 has this property, so C2C_2 is the director circle of C1C_1. So (4) is correct. Step 3: OPOP bisects ∠APB\angle APB, so ∠OPM=45∘\angle OPM = 45^\circ, where MM is the point of contact on PAPA. In the right triangle OMPOMP, right angled at MM since OM⊥PAOM \perp PA,
sin⁡45∘=OMOP=r1r2⟹12=r1r2\sin 45^\circ = \frac{OM}{OP} = \frac{r_1}{r_2} \quad\Longrightarrow\quad \frac{1}{\sqrt2} = \frac{r_1}{r_2}
r2=2 r1r_2 = \sqrt2\, r_1
 So (2) is correct. Step 4: (1) needs r2=2r1r_2=2r_1, which contradicts r2=2r1r_2=\sqrt2 r_1, so it is false. (3) claims r22<2r12r_2^2 < 2r_1^2, but r22=2r12r_2^2 = 2r_1^2 exactly, so the strict inequality fails. Answer: (2) and (4)
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