JEE Main 2025 Apr 4 Shift 1, Mathematics Q23: Involving Two or More Circles

JEE Main2025Apr 4, Shift 1Mathematics

Let $A = {z \in C : ∣ z - 2 - i ∣ = 3}$ , $B = {z \in C : Re (z - i z) = 2}$ and $S = A \cap B$ . Then $z \in S \sum ∣ z ∣^{2}$ is equal to ____.

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Solution

Write $z = x + i y$ . Then $A$ is the circle $(x - 2)^{2} + (y - 1)^{2} = 9$ .

For $B$ : $z - i z = (x + i y) - i (x + i y) = (x + y) + i (y - x)$ . So $Re (z - i z) = x + y = 2$ .

Substitute $y = 2 - x$ in the circle: $(x - 2)^{2} + (1 - x)^{2} = 9 \Rightarrow 2 x^{2} - 6 x - 4 = 0 \Rightarrow x^{2} - 3 x - 2 = 0$ , giving $x = \frac{3 \pm 17}{2}$ and correspondingly $y = \frac{1 \mp 17}{2}$ .

For each root, $∣ z ∣^{2} = x^{2} + y^{2}$ . Using $x + y = 2$ and the sum/product from the quadratics:

$x^{2} + y^{2} = (x + y)^{2} - 2 x y = 4 - 2 x y$ . With $x = \frac{3 + 17}{2}$ and $y = \frac{1 - 17}{2}$ , $x y = \frac{3 - 3 17 + 17 - 17}{4} = \frac{- 14 - 2 17}{4} = \frac{- 7 - 17}{2}$ .

So $∣ z_{1} ∣^{2} = 4 - 2 \cdot \frac{- 7 - 17}{2} = 11 + 17$ , and by symmetry $∣ z_{2} ∣^{2} = 11 - 17$ .

Sum $= 22$ .

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JEE Main 2025 Apr 4 Shift 1, Mathematics Q23: Involving Two or More Circles

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