Let |(z̄ − i)/(2z̄ + i)| = 1/3, z ∈ C, Be the Equation of a Circle With Centre C. If the Area of the Triangle Whose Vertices Are (0, 0), C and (α, 0) Is 11 Square Units, Then α² Equals

JEE Main 2025 — 23 January, Shift 1. Previous Year Question.

Problem

Let $\left|\dfrac{\bar z-i}{2\bar z+i}\right|=\dfrac13$, $z\in\mathbb C$, be the equation of a circle with centre $C$. If the area of the triangle whose vertices are the points $(0,0)$, $C$ and $(\alpha,0)$ is $11$ square units, then $\alpha^2$ equals:

Note: the title and description attached to the worked solution embedded below describe a different problem. The question actually solved there, from start to finish, is this one.

Key insight. Writing $z=x+iy$ turns the modulus condition into an ordinary algebraic equation in $x$ and $y$ once both sides are squared — the messy-looking ratio of conjugates is really just a circle in disguise, and reading off its centre is all the geometry that follows actually needs.

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Approach

Substitute $z=x+iy$ (so $\bar z=x-iy$) into the modulus equation and square both sides to clear the square roots. Cross-multiply and simplify into the standard form of a circle to read off its centre $C$. The triangle formed by the origin, $C$ and $(\alpha,0)$ turns out to be right-angled at the origin, so its area is half the product of the two legs — one leg is $|\alpha|$ along the $x$-axis, the other is the $y$-coordinate of $C$.

Solution

Step 1 — Substitute z = x + iy

$$\bar z-i = x-iy-i = x-i(y+1), \qquad 2\bar z+i = 2x-2iy+i = 2x+i(1-2y)$$

Step 2 — Write the modulus condition and square both sides

$$\frac{\sqrt{x^2+(y+1)^2}}{\sqrt{4x^2+(1-2y)^2}} = \frac13 \implies 9\left[x^2+(y+1)^2\right] = 4x^2+(1-2y)^2$$

Step 3 — Expand both sides

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

Step 4 — Collect terms into the equation of a circle

$$5x^2+5y^2+22y+8=0 \implies x^2+y^2+\frac{22}5y+\frac85=0$$

Step 5 — Read off the centre

Comparing with $x^2+y^2+2gx+2fy+c=0$, the centre is $(-g,-f)$, so here:

$$C = \left(0,-\frac{11}5\right)$$

Step 6 — Set up the triangle’s area

The three vertices are $(0,0)$, $C=\left(0,-\dfrac{11}5\right)$ and $(\alpha,0)$. Since the origin-to-$C$ segment lies entirely on the $y$-axis and the origin-to-$(\alpha,0)$ segment lies entirely on the $x$-axis, the triangle is right-angled at the origin, with legs of length $\dfrac{11}5$ and $|\alpha|$.

Step 7 — Solve for α²

$$\text{Area} = \frac12\cdot|\alpha|\cdot\frac{11}5 = 11 \implies |\alpha| = 10 \implies \alpha^2 = 100$$

Answer

Common mistakes

• Squaring the modulus equation before fully simplifying $\bar z-i$ and $2\bar z+i$ into real-and-imaginary-part form. Trying to square the fraction directly, without first separating each side into $x$- and $y$-terms, tangles the algebra unnecessarily — substitute $z=x+iy$ first, then square.
• Sign errors when expanding $(1-2y)^2$ or $(y+1)^2$. Both expansions carry a cross term (respectively $-4y$ and $+2y$) that is easy to drop, and a dropped cross term shifts the circle’s centre.
• Forgetting that the area formula needs $|\alpha|$, not $\alpha$. The triangle’s area is a positive quantity, so the equation $\frac12\cdot|\alpha|\cdot\frac{11}5=11$ gives $|\alpha|=10$ directly — squaring at the end removes the sign ambiguity, but it’s worth noticing the problem never actually needs the sign of $\alpha$ itself.

Practise next

• Let $\left|\dfrac{\bar z-3i}{\bar z+3i}\right|=\dfrac12$, $z\in\mathbb C$, be the equation of a circle with centre $C$. If the area of the triangle whose vertices are $(0,0)$, $C$ and $(\beta,0)$ is $20$ square units, find $\beta^2$.