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# Prove that the equation of circle in the$z$plane can be written in the form$\alpha z\overline z + \overline \beta z + \beta \overline z + c = 0$. Deduce the equation of the line.A. $\overline \beta z + \beta \overline z + c = 0$B. $\overline \beta z - \beta \overline z + c = 0$C. $\overline \beta z + \beta \overline z - c = 0$D. None of these  Answer Verified
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$\overline z = x - iy$and$x = \dfrac{{z + \overline z }}{2},y = \dfrac{{z - \overline z }}{{2i}},z\overline z = |z{|^2} = {x^2} + {y^2}$. 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:
$\alpha ({x^2} + {y^2}) + 2gx + 2fy + c = 0 \\ \Rightarrow \alpha (z\overline z ) + g(z + \overline z ) + f(\dfrac{{z - \overline z }}{i}) + c = 0{\text{ }}[{x^2} + {y^2} = z\overline z ,\dfrac{{z + \overline z }}{2} = x,\dfrac{{z - \overline z }}{{2i}} = y] \\ \Rightarrow \alpha (z\overline z ) + g(z + \overline z ) - if(z - \overline z ) + c = 0 \\ \Rightarrow \alpha (z\overline z ) + (g - if)z + (g + if)\overline z + c = 0 \\ \Rightarrow \alpha (z\overline z ) + (\overline \beta )z + (\beta )\overline z + c = 0{\text{ }}[\overline \beta = g - if,\beta = g + if] \\ \Rightarrow \alpha z\overline z + \overline \beta z + \beta \overline z + c = 0 \\$
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$\overline \beta z + \beta \overline z + c = 0$. 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.
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