Question

In the figure $|z|=r$ is circumcircle of $\Delta ABC.D,E$ & F are the middle points of the sides BC, CA & AB respectively, AD produced to meet the circle at L. If $\angle CAD=\theta ,AD=x,BD=y$ and altitude of $\Delta ABC$ from A meet the circle $|z|=r$ at $M,z_a,z_b$ & $z_c$ are affixes of vertices A, B & C respectively.

Affix of M is

• A. $2z_b{e}^{i2B}$
• B. $z_b{e}^{i(\pi -2B)}$
• C. $z_b{e}^{iB}$
• D. $2z_b{e}^{iB}$

Answer 1

Answer: (B)

$z_b{e}^{i(\pi -2B)}$

From the figure $\left|z\right|=r,$ circumcircle of $△ABC.$

Given $\angle CAD=\theta ,$ $AD=x$ and $BD=y$

Circle $\left|z\right|=r,$ meet at $M$

$\Rightarrow \left|z\right|=r$ implies that , $z=e^{ri\theta },\theta \in R$

$\Rightarrow \frac{2}{zB}=\frac{2}{e^{ri\theta }}=zb{e}^{ri\theta }$

$\Rightarrow r\theta =(\pi -2B)$

$\Rightarrow$ If $\theta \in R=-\theta \in R$

$\therefore z_b{e}^{\pi -2B}$ is a circle.

Hence, the answer is $z_b{e}^{\pi -2B}.$