Question

If $A\left(z_1\right)$ is a variable point in the Argand plane such that $z_1\overline{z_1}=5,0<\arg \left(z_1\right)<\pi$. Also $B\left(z_2\right)\&C\left(z_3\right)$ are two fixed points in the Argand plane satisfying $z^4-6z^2+25=0\&-\pi <arg\left(z\right)<0$.

• A. Internal angle bisector of $\angle A$ of $△ABC$ will always pass through a fixed point.
• B. Circum centre of $△ABC$ is the Origin.
• C. Orthocentre $\left(z_4\right)$ of $△ABC$ satisfies $|z_4+2i|=5.(i=\sqrt{-1})$
• D. Orthocentre $\left(z_4\right)$ of $△ABC$ satisfies $|z_4-2i|=\sqrt{5}$.

Answer 1

Answer: (A), (B)

The correct options are
A Internal angle bisector of $\angle A$ of $△ABC$ will always pass through a fixed point.
B Circum centre of $△ABC$ is the Origin.

$|z_1|=\sqrt{5},z_2,z_3=\pm 2-i$

$|z_1|=|z_2|=|z_3|=\sqrt{5}$

So, $z_0$(circum centre)$=0+0.i$ (Origin)

Angle bisector of A passes through mid point of minor $\text{arc}\left(BC\right)=D$.

Let $z_4(\text{Orthocenter)}=x+iy,z_1=\sqrt{5}\cos \theta +i\sqrt{5}\sin \theta ,z_2=2-i,z_3=-2-i$

$\left(O\right(\text{Circumcenter})=0,G(\text{Centroid}=\sqrt{5}\cos \theta +i(\sqrt{5}\sin \theta -2))$

We know that the centroid devides the line joining the orthocenter & the circumcenter into 2:1 internally.
By solving, we get $|z_4+2i|=\sqrt{5}$