Let w = z z + k 1 z + k 2 i z + ( 1 + i ) , k 1 , k 2 . Let Re ( w ) = 0 be the circle C of radius 1 in the first quadrant touching the line y = 1 and the y-axis. If the curve Im ( w ) = 0 intersects C at A and B, then 30 ( A B ) 2 is

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Solution

Q.
$Let w = z \bar{z} + k_{1} z + k_{2} i z + λ (1 + i), k_{1}, k_{2} \in ℝ .$ $Let Re (w) = 0 be the circle C of radius 1$ in the first quadrant touching the line $y = 1$ and the y-axis. If the curve $Im (w) = 0$ intersects C at A and B, then $30 {(A B)}^{2}$ is equal to ______. [2023]

Ans.
(24)

We have, $w = z \bar{z} + k_{1} z + k_{2} i z + λ (1 + i)$

Put $z = x + i y$

$w = x^{2} + y^{2} + k_{1} (x + i y) + k_{2} i (x + i y) + λ (1 + i)$

$= x^{2} + y^{2} + k_{1} x + k_{1} i y + k_{2} i x - k_{2} y + λ + λ i$

$= (x^{2} + y^{2} + k_{1} x - k_{2} y + λ) + i (k_{1} y + k_{2} x + λ)$

Given $Re (w) = 0$

$\Rightarrow x^{2} + y^{2} + k_{1} x - k_{2} y + λ = 0$ $...(i)$

$Centre = (\frac{- k_{1}}{2}, \frac{k_{2}}{2})$

$Radius = \sqrt{\frac{k_{1}^{2}}{4} + \frac{k_{2}^{2}}{4} - λ}$

Now, $\frac{- k_{1}}{2} = 1 \Rightarrow k_{1} = - 2$

and $\frac{k_{2}}{2} = 2 \Rightarrow k_{2} = 4$

Also, $\sqrt{\frac{k_{1}^{2}}{4} + \frac{k_{2}^{2}}{4} - λ} = 1$

$\Rightarrow \frac{k_{1}^{2}}{4} + \frac{k_{2}^{2}}{4} - λ = 1$

$\Rightarrow 1 + 4 - λ = 1 \Rightarrow λ = 4$

Put $k_{1} = - 2, k_{2} = 4, λ = 4$ in (i)

$x^{2} + y^{2} - 2 x - 4 y + 4 = 0$ $...(ii)$

Now, $Im (w) = 0,$ $k_{2} x + k_{1} y + λ = 0$

$4 x - 2 y + 4 = 0$ $...(iii)$

From (ii) and (iii), $A \equiv (0, 2)$ and $B \equiv (0.4, 2.8)$

Now, $A B = \sqrt{{(0.4 - 0)}^{2} + {(2.8 - 2)}^{2}}$

$\Rightarrow A B^{2} = {(0.4)}^{2} + {(0.8)}^{2} = 0.8$

$∴$ $30 {(A B)}^{2} = 30 (0.8) = 24$

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