Let |z−i|/|2z+i| = 1/3, z ∈ C, Be the Equation of a Circle With Centre at 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 $\dfrac{|z-i|}{|2z+i|}=\dfrac13$, $z\in\mathbb C$, be the equation of a circle with center at $C$. If the area of the triangle whose vertices are at the points $(0,0)$, $C$ and $(\alpha,0)$ is $11$ square units, then $\alpha^2$ equals:

Key insight. Squaring the given modulus ratio and substituting $z=x+iy$ turns the condition into an ordinary Cartesian circle equation — from there, the triangle with two vertices on the axes has an area that’s just half the base times the height, with no need for the circle’s radius at all.

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Approach

Substitute $z=x+iy$ into the given modulus condition and clear the square roots by squaring both sides, which turns the equation into a standard circle equation in $x$ and $y$. Read off the center $C$ by comparing with the general form $x^2+y^2+Dx+Ey+F=0$. Since the triangle has vertices at the origin, at $C$ on the $y$-axis, and at $(\alpha,0)$ on the $x$-axis, its area is simply half the product of the two legs along the axes — solve that for $\alpha^2$.

Solution

Step 1 — Set up the condition with z = x + iy

$$\frac{|z-i|}{|2z+i|} = \frac13 \implies 3|z-i| = |2z+i| \implies 9|z-i|^2 = |2z+i|^2$$

With $z=x+iy$:

$$|z-i|^2 = x^2+(y-1)^2, \qquad |2z+i|^2 = (2x)^2+(2y+1)^2 = 4x^2+(2y+1)^2$$

Step 2 — Expand both sides

$$9\left[x^2+(y-1)^2\right] = 4x^2+(2y+1)^2$$

$$9x^2+9(y^2-2y+1) = 4x^2+4y^2+4y+1$$

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

Step 3 — Collect into standard circle form

Bringing everything to one side:

$$5x^2+5y^2-22y+8 = 0$$

Dividing by 5:

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

Step 4 — Read off the center C

Comparing with the general circle equation $x^2+y^2+Dx+Ey+F=0$, whose center is $\left(-\dfrac D2,-\dfrac E2\right)$: here $D=0$ and $E=-\dfrac{22}5$, so

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

Step 5 — Set up the triangle’s area

The triangle has vertices at the origin $(0,0)$, at $C=\left(0,\frac{11}5\right)$ on the $y$-axis, and at $(\alpha,0)$ on the $x$-axis. Since two of its sides lie along the coordinate axes, it’s a right triangle with legs $|\alpha|$ (along the $x$-axis) and $\dfrac{11}5$ (along the $y$-axis, the perpendicular distance from $C$ to the origin):

$$\text{Area} = \frac12\times|\alpha|\times\frac{11}5 = 11$$

Step 6 — Solve for α, then α²

$$|\alpha|\times\frac{11}5 = 22 \implies |\alpha| = 22\times\frac5{11} = 10$$

$$\alpha^2 = 10^2 = 100$$

Answer

Common mistakes

• Forgetting to square the modulus ratio before clearing denominators. Working with $|z-i|=\frac13|2z+i|$ directly still involves square roots buried inside each modulus — squaring both sides first, as in Step 1, is what turns this into ordinary algebra.
• Sign or arithmetic slips while collecting the x² and y² terms. With coefficients like $9x^2-4x^2$ and $9y^2-4y^2$ on either side, it’s easy to combine them incorrectly — double-checking against the standard form $x^2+y^2+Dx+Ey+F=0$ (with a leading coefficient of exactly 1 on $x^2$ and $y^2$) catches this.
• Not recognizing that the triangle’s area formula needs no distance formula here. Because two vertices sit exactly on the axes and the third is the origin, the area is immediately $\frac12(\text{base})(\text{height})$ — reaching for the general three-point area formula is unnecessary extra work.

Practise next

• Let $\dfrac{|z-2i|}{|z+2i|}=\dfrac12$ be the equation of a circle with center $C$. If the triangle with vertices $(0,0)$, $C$ and $(\beta,0)$ has area $6$, find $\beta^2$ using the same approach.