QJEEM23Ma78bb: JEE Main 2023 Mathematics - Class 11 JEEM23Ma78bb Solution: Let w = z z ¯ + k 1 z + k 2 i z + λ ( 1 + i ) , k 1 , k 2 ∈ R . …

Given the expression for $w$ as :

$$w=z\overline{z}+k_{1}z+k_{2}iz+\lambda (1+i),\text{where }{k}_1,k_2\in \mathbb{R}.$$

1. If $w = x+iy$ , we can separate this into the real and imaginary parts :

The real part is: $$\text{Re}(w)=x^2+y^2+k_{1}x-k_{2}y+\lambda =0.$$

The imaginary part is: $$\text{Im}(w)=k_{1}y+k_{2}x+\lambda =0.$$

2. We know the circle C of radius 1 touches the line $y=1$ and the y-axis. The standard form of the equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$ , where $(h, k)$ is the center of the circle and $r$ is the radius. If the circle touches y-axis, then the x-coordinate of the center $h$ must be equal to the radius. Also, the circle touches the line $y=1$ , so the y-coordinate of the center $k$ must be 1 unit above this line, hence $k=2$ . So the circle's equation is $(x-1)^2 + (y-2)^2 = 1$ .

3. By comparing this with $\text{Re}(w) = 0$ , we can see that $k_1=-2$ , $k_2=4$ , and $\lambda=4$ .

4. The equation for the curve $\text{Im}(w)=0$ becomes $-2y + 4x + 4 = 0$ , which simplifies to $y = 2x + 2$ .

5. To find the intersection points A and B, we solve the system of equations consisting of the circle and the line. Substituting $y = 2x + 2$ into the equation of the circle gives $5x^2 - 2x = 0$ , with solutions $x = 0$ and $x = 2/5$ .

6. Substituting $x = 0$ and $x = 2/5$ into $y = 2x + 2$ gives $y = 2$ and $y = 6/5$ respectively. So, the intersection points are A = $(0, 2)$ and B = $(2/5, 14/5)$ .

7. The distance between A and B is $AB = \sqrt{[(2/5)^2 + (4/5)^2]} = \sqrt{4/5}$ .

8. Finally, $30(AB)^2 = 30 \times (4/5) = 24$ .