JEE Main 2021 — Circle Question with Solution

S₁ : x² + y² = 9 ; C₁ (0, 0), r₁ = 3

S₂ : (x $$-$$ 2)² + y² = 1 ; C₂ (2, 0), r₂ = 1

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Let the variable circle S and its radius is r units.

Here S and S₁ touches internally

$$\therefore$$ Distance between center,

S + S₁ = PC₁ = 3 $$-$$ r

Here S and S₂ touches externally

$$\therefore$$ Distance between center,

S + S₂ = PC₂ = 1 + r

$$\therefore$$ PC₁ + PC₂ = 4 > C₁ C₂

So locus is ellipse whose focii are C₁ & C₂ and major axis is 2a = 4 and 2ae = C₁C₂ = 2

$$\Rightarrow$$ $$e=\frac{1}{2}$$

$$\Rightarrow$$ $$b^2=4\left(1-\frac{1}{4}\right)=3$$

Centre of ellipse is midpoint of C₁ & C₂ is (1, 0)

Equation of ellipse is $$\frac{{(x-1)}^2}{2^2}+\frac{y^2}{{\left(\sqrt{3}\right)}^2}=1$$

Now by cross checking the option $$\left(2,\pm \frac{3}{2}\right)$$ satisfied it.