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

 

 

 

 

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