## Tag Archives: concyclic points

### Equation of a circle

Consider a fixed complex number $z_{0}$ and let $z$ be any complex number which moves in such a way that its distance from $z_{0}$ is always equal to r. This implies $z$ would lie on a circle whose centre is $z_{0}$ and radius r. And, its equation would be

$|z-z_{0}|=r$

or $|z-z_{0}|^{2}=r^{2}$

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

or $z\overline{z}-z \overline{z_{0}}-\overline{z}z_{0}-r^{2}=0$

Let $-a=z_{0}$ 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 line segment AB $(A(z_{1}), B(z_{2}))$ as its diameter. Let $P(z)$ as its diameter. Let $P(z)$ be any point on the circle. As the angle in the semicircle is $\pi/2$, so

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

$\Longrightarrow (\frac{z_{1}-z}{z_{2}-z})=\pm \pi/2$

$\Longrightarrow \frac{z-z_{1}}{z-z_{2}}$ is purely imaginary.

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

$\Longrightarrow (z-z_{1})(\overline{z}-\overline{z_{2}})+(z-z_{2})(\overline{z}-\overline{z_{1}})=0$

Condition for four points to be concyclic:

Let ABCD be a cyclic quadrilateral such that $A(z_{1})$, $B(z_{2})$, $C(z_{3})$ and $D(z_{4})$ lie on a circle. (Remember the following basic property of concyclic quadrilaterals: opposite angles are supplementary).

The above property means the following:

$\arg (\frac{z_{4}-z_{1}}{z_{2}-z_{1}})+\arg (\frac{z_{2}-z_{3}}{z_{4}-z_{3}})=\pi$

$\Longrightarrow \arg (\frac{z_{4}-z_{1}}{z_{2}-z_{1}})(\frac{z_{2}-z_{3}}{z_{4}-z_{3}})=\pi$

$(\frac{z_{4}-z_{1}}{z_{2}-z_{1}})(\frac{z_{2}-z_{3}}{z_{4}-z_{3}})$ is purely real.

Thus, points $A(z_{1})$, $B(z_{2})$, $C(z_{3})$, $D(z_{4})$ (taken in order) would be concyclic if the above condition is satisfied.

More later,

Nalin Pithwa