# Complex Numbers – Equation of a Circle

## Complex Numbers – Equation of a Circle

Equation of a Circle: Consider a fixed complex number zₒ and let z be any complex number which moves in such a way that its distance from zₒ is always to r. this implies z would lie on a circle whose center is zₒ and radius is r. and its equation would be

|z – zₒ| = r

Squaring on both sides

|z – zₒ|² = r²

$$(z-{{z}_{0}})(\overline{z}-\overline{{{z}_{0}}})={{r}^{2}}$$,

$$z\overline{z}-z\overline{{{z}_{0}}}-\overline{z}{{z}_{0}}+{{z}_{0}}\overline{{{z}_{0}}}-{{r}^{2}}=0$$.

Let -a = zₒ and $${{z}_{0}}\overline{{{z}_{0}}}-{{r}^{2}}=b$$. Then

$$z\overline{z}+a\overline{z}+\overline{a}z+b=0$$.

It represents the general equation of a circle in the complex plane.

Now, let us consider a circle described on a lime segment AB $$(A({{z}_{1}}),B({{z}_{2}}))$$ as its diameter. Let P(z) be any point on the circle. As the angle in the semicircle is $$\frac{\pi }{2}$$, So

$$\angle APB=\frac{\pi }{2}$$

$$\arg \left( \frac{{{z}_{1}}-z}{{{z}_{2}}-z} \right)=\pm \frac{\pi }{2}$$,

$$\frac{z-{{z}_{1}}}{z-{{z}_{2}}}$$ is purely imaginary

$$\frac{z-{{z}_{1}}}{z-{{z}_{2}}}+\frac{\overline{z}-\overline{{{z}_{1}}}}{\overline{z}-{{\overline{z}}_{2}}}=0$$,

$$(z-{{z}_{1}})(\overline{z}-\overline{{{z}_{2}}})+(z-{{z}_{2}})(\overline{z}-\overline{{{z}_{1}}})=0$$.