Question

Let $S=\{z \mid z \bar{z}-(3-4 i) z-(3+4 i) \bar{z}+21=0\}$. 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

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

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