Question

Let $S=\{z\mid z\bar{z}-(3-4i)z-(3+4i)\bar{z}+21=0\}$. If $M$ and $m$ be the maximum and minimum value of $\left(\frac{z-\bar{z}}{i(z+\bar{z})}\right)$ then find $\left(\frac{1}{M}+\frac{1}{m}\right)$

Answer 1

Answer: 2

$z\mid z\bar{z}-(3-4i)z-(3+4i)\bar{z}+21=0$
$z\overline{z}+a\overline{z}+\overline{a}z+b=0\left(\mathrm{Centre}(-a),r=\sqrt{|a{|}^2-b}\right)$
Radius $\sqrt{9+16-21}=\sqrt{4}=2$
Centre $\to (3,4)$
Circle $(x-3{)}^2+(y-4{)}^2=4$
$\frac{z-\overline{z}}{i(z+\overline{z})}\Rightarrow \frac{2\times 2iy}{i(2x)}\Rightarrow \left(\frac{y}{x}\right)$
Equation of tangent
$(y-4)=m(x-3)\pm 2\sqrt{1+m^2}$
Satisfies $(0,0)$

$3m-4=\pm 2\sqrt{1+m^2}\Rightarrow 9m^2+16-24m=2\left(1+m^2\right)\Rightarrow 5m^2-24m+12=0$
$\frac{1}{M}+\frac{1}{m}=\left(\frac{M+m}{Mm}\right)\Rightarrow \frac{24}{5}\times \frac{5}{12}=2$