JEE Main 2025 Jan 23 Shift 1, Mathematics Q10: Introduction to Complex Numbers And Their Conjugates

JEE Main2025Jan 23, Shift 1Mathematics

Let $\frac{z ˉ - i}{2 z ˉ + i} = \frac{1}{3}$ , $z \in C$ , be the equation of a circle with centre at C. If the area of the triangle, whose vertices are at the points $(0, 0)$ , $C$ , and $(α, 0)$ is 11 square units, then $α^{2}$ equals

A
$100$
B
$50$
C
$\frac{121}{25}$
D
$\frac{81}{25}$

Solution

Write $z = x + i y$ , so $\overset{z}{ˉ} = x - i y$ . The condition becomes $∣ \overset{z}{ˉ} - i ∣ = \frac{1}{3} ∣2 \overset{z}{ˉ} + i ∣$ , i.e. $3∣ \overset{z}{ˉ} - i ∣ = ∣2 \overset{z}{ˉ} + i ∣$ .

$3∣ x - i y - i ∣ = ∣2 x - 2 i y + i ∣$ , so $9 [x^{2} + (y + 1)^{2}] = 4 x^{2} + (1 - 2 y)^{2}$ .

$9 x^{2} + 9 y^{2} + 18 y + 9 = 4 x^{2} + 4 y^{2} - 4 y + 1$

$\Rightarrow 5 x^{2} + 5 y^{2} + 22 y + 8 = 0 \Rightarrow x^{2} + y^{2} + \frac{22}{5} y + \frac{8}{5} = 0$ .

Centre $C = (0, - \frac{11}{5})$ .

Area of triangle with vertices $(0, 0)$ , $C = (0, - 11/5)$ , $(α, 0)$ is $\frac{1}{2} ∣ α ∣ \cdot \frac{11}{5} = 11$ .

$∣ α ∣ = 10 \Rightarrow α^{2} = 100$ .

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JEE Main 2025 Jan 23 Shift 1, Mathematics Q10: Introduction to Complex Numbers And Their Conjugates

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