Let = z z + k1z + k2iz + (1+i), k1, k2 R. Let Re() = 0 be the cir… — Solution | JEE Mathematics

NVAMediumJEE 2023Complex Numbers Basics

Let $\omega = z\bar{z} + k_1z + k_2iz + \lambda(1+i)$ , $k_1, k_2 \in \mathbb{R}$ . Let $Re(\omega) = 0$ be the circle $C$ of radius $1$ in the first quadrant touching the line $y = 1$ and the y-axis. If the curve $\text{Im}(\omega) = 0$ intersects $C$ at $A$ and $B$ , then $30(AB)^2$ is equal to _____

Answer

Correct answer:24

Step-by-step solution

Standard Method

Given:

\omega = z\bar{z} + k_1z + k_2\bar{z} + \lambda(1+i)

with $k_1, k_2 \in \mathbb{R}$ .

Find: $30(AB)^2$ where the curve $\text{Im}(\omega)=0$ intersects the circle $C$ given by $\text{Re}(\omega)=0$ .

From the solution, the real part is

\text{Re}(\omega) = x^2 + y^2 + k_1x + k_2y + \lambda = 0

Comparing with the general circle

x^2 + y^2 + 2gx + 2fy + c = 0

its center is

\left(-\frac{k_1}{2}, -\frac{k_2}{2}\right)

The circle has radius $1$ , lies in the first quadrant, touches the y-axis and the line $y=1$ . Hence the center is $(1,2)$ , so

-\frac{k_1}{2} = 1, \quad -\frac{k_2}{2} = 2

Therefore,

k_1 = -2, \quad k_2 = -4, \quad \lambda = 4

Intersection and chord length

Now the imaginary part is

\text{Im}(\omega) = k_1y - k_2x + \lambda = 0

Substituting $k_1=-2$ , $k_2=-4$ and $\lambda=4$ ,

-2y - (-4)x + 4 = 0

2x - y + 2 = 0

y = 2x + 2

Compute $$30(AB)^2$$

The circle equation becomes

x^2 + y^2 - 2x - 4y + 4 = 0

Substitute $y=2x+2$ :

x^2 + (2x+2)^2 - 2x - 4(2x+2) + 4 = 0

x^2 + 4x^2 + 8x + 4 - 2x - 8x - 8 + 4 = 0

5x^2 - 2x = 0

Thus the intersection points are $A=(0,2)$ and $B=\left(\frac{2}{5}, \frac{14}{5}\right)$ . Then

AB = \sqrt{\left(\frac{2}{5}\right)^2 + \left(\frac{4}{5}\right)^2} = \sqrt{\frac{20}{25}} = \frac{2\sqrt{5}}{5}

Hence

(AB)^2 = \frac{4}{5}

and

30(AB)^2 = 30 \cdot \frac{4}{5} = 24

Therefore, the final answer is $24$ .

Common mistakes

Using the wrong form of $\omega$ . The solution works with $k_2\bar{z}$ , not $k_2iz$ . If this is copied incorrectly, the real and imaginary parts change. Always split $\omega$ into real and imaginary parts exactly as used in the solution.
Misreading the tangency conditions. A circle touching the y-axis and the line $y=1$ with radius $1$ has center at horizontal distance $1$ from the y-axis and vertical distance $1$ from the line $y=1$ . Use these distances to identify the center correctly.
Making an algebra sign error while substituting the line into the circle. This changes the intersection points and hence the chord length. Substitute carefully and simplify term by term.

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