JEE Mathematics 2025 Question with Solution

Given: $\left| \frac{z-i}{2z+i} \right| = \frac{1}{3}$ is used in the extracted working to obtain the circle, and the area of the triangle formed by $(0,0)$, the center $C$, and $(\alpha,0)$ is $11$ square units.

Find: $\alpha^2$.

From the solution working,

$$\left| \frac{z-i}{z+\frac{i}{2}} \right| = \frac{2}{3}$$

Taking $z=x+iy$, this gives

$$3\left|x+i(y-1)\right|=2\left|x+i\left(y+\frac{1}{2}\right)\right|$$

So,

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

Squaring and simplifying as shown in the extracted solution,

$$9\left(x^2+(y+1)^2\right)=4\left(x^2+\left(y-\frac{1}{2}\right)^2\right)$$

which leads to

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

Dividing by $5$,

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

Hence the center of the circle is

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

The base of the triangle lies on the $x$-axis from $(0,0)$ to $(\alpha,0)$, so its length is $|\alpha|$. The perpendicular distance of the center from the $x$-axis is

$$\left| -\frac{11}{5} \right|=\frac{11}{5}$$

Therefore the area is

$$\frac{1}{2}\cdot |\alpha| \cdot \frac{11}{5}=11$$

So,

$$|\alpha|\cdot \frac{11}{10}=11$$ $$|\alpha|=10$$

Thus,

$$\alpha^2=100$$

Therefore, the correct option is A.

Note: The first extracted approach contains inconsistent algebra and an intermediate value $\alpha^2=484$, but the second approach correctly yields the center and the final answer $\alpha^2=100$.

Given: The circle obtained from the provided working has center $C=\left(0,-\frac{11}{5}\right)$.

Find: The value of $\alpha^2$ using the triangle area.

The three vertices are

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

Since $A$ and $B$ lie on the $x$-axis, the line through them is

$$y=0$$

So the height from $C$ to the base $AB$ is

$$\frac{11}{5}$$

The base length is

$$AB=|\alpha|$$

Using the triangle area formula,

$$\text{Area}=\frac{1}{2}\times \text{base} \times \text{height}$$

Therefore,

$$11=\frac{1}{2}\times |\alpha| \times \frac{11}{5}$$

Multiply both sides by $\frac{10}{11}$:

$$|\alpha|=10$$

Hence,

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

Therefore, the required value is $100$.