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  4. Let two curves C1: x2+y2=2 and C2: locus of z which satisfies ||z+3 √2|-| z-3 √2||=2 √2. If locus of z satisfying | arg ( z -1)|= tan -1(4) meets the curve C 2 at A and B then area of the triangle formed by A , B and C where complex number corresponding to C is e i 2 π, is

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Q. Let two curves $C_1: x^2+y^2=2$ and $C_2:$ locus of $z$ which satisfies ||$z+3 \sqrt{2}|-| z-3 \sqrt{2}||=2 \sqrt{2}$.
If locus of $z$ satisfying $|\arg ( z -1)|=\tan ^{-1}(4)$ meets the curve $C _2$ at $A$ and $B$ then area of the triangle formed by $A , B$ and $C$ where complex number corresponding to $C$ is $e ^{ i 2 \pi}$, is

Complex Numbers and Quadratic Equations

Solution:

Using triangle inequality
$\left|\left(z^2-2 i z\right)-\left(z^2-9 z+7 i z\right)\right| \leq 18 $
$|9 z-9 i z| \leq 18 \Rightarrow|z| \leq \sqrt{2}$
$C_1: x^2+y^2=2$
$C_2: \frac{x^2}{2}-\frac{y^2}{16}=1 $
image
Line $A B$
$y=4 x-4$
which is tangent to thecurve $C_2$.
$A \equiv(2,4), B \equiv(2,-4) \text { and } C \equiv(1,0) $
$\therefore \operatorname{Ar}(\triangle ABC )=\frac{1}{2} \times 1 \times 8=4 \text { sq. units. }$