Question

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 $e^{i2\pi }$, is

• A. 4 sq. units
• B. $8\sqrt{2}$ sq. units
• C. 16 sq. units
• D. $16\sqrt{2}$ sq. units

Answer 1

Answer: (A)

Using triangle inequality
$\left|\left(z^2-2iz\right)-\left(z^2-9z+7iz\right)\right|\le 18$
$|9z-9iz|\le 18\Rightarrow |z|\le \sqrt{2}$
$C_1:x^2+y^2=2$
$C_2:\frac{x^2}{2}-\frac{y^2}{16}=1$

Line $AB$
$y=4x-4$
which is tangent to thecurve $C_2$.
$A\equiv (2,4),B\equiv (2,-4)\text{ and }C\equiv (1,0)$
$\therefore \mathrm{Ar}(△ABC)=\frac{1}{2}\times 1\times 8=4\text{ sq. units. }$