Question

Prove that the equation of circle in the$z$plane can be written in the form$\alpha z\bar{z}+\bar{\beta }z+\beta \bar{z}+c=0$. Deduce the equation of the line.
A. $\bar{\beta }z+\beta \bar{z}+c=0$
B. $\bar{\beta }z-\beta \bar{z}+c=0$
C. $\bar{\beta }z+\beta \bar{z}-c=0$
D. None of these

Answer 1

Hint: Consider the standard form of circle in coordinate geometry then use basic formulas of complex numbers to convert it into complex form.

We know that, if$z=x+iy$then$\bar{z}=x-iy$and$x={\displaystyle \frac{z+\bar{z}}{2}},y={\displaystyle \frac{z-\bar{z}}{2i}},z\bar{z}=|z{|}^2=x^2+y^2$. The standard equation of the circle is$\alpha (x^2+y^2)+2gx+2fy+c=0$.We’ll use above mentioned formula to solve further as follows:
 $$\begin{gathered} \alpha (x^2+y^2)+2gx+2fy+c=0 \\ \Rightarrow \alpha (z\bar{z})+g(z+\bar{z})+f({\displaystyle \frac{z-\bar{z}}{i}})+c=0[x^2+y^2=z\bar{z},{\displaystyle \frac{z+\bar{z}}{2}}=x,{\displaystyle \frac{z-\bar{z}}{2i}}=y] \\ \Rightarrow \alpha (z\bar{z})+g(z+\bar{z})-if(z-\bar{z})+c=0 \\ \Rightarrow \alpha (z\bar{z})+(g-if)z+(g+if)\bar{z}+c=0 \\ \Rightarrow \alpha (z\bar{z})+(\bar{\beta })z+(\beta )\bar{z}+c=0[\bar{\beta }=g-if,\beta =g+if] \\ \Rightarrow \alpha z\bar{z}+\bar{\beta }z+\beta \bar{z}+c=0 \end{gathered}$$
It is in the same form as the given equation. Now observe from the standard form of the circle that if we put$\alpha =0$then we’ll get the equation of a straight line. Hence putting$\alpha =0$in the given equation we’ll get$\bar{\beta }z+\beta \bar{z}+c=0$. Hence option A is the correct option.

Note: The hack in this question was to observe that, what’s the relation between the equation of a circle and straight line in the coordinate plane.