Show that the equation of a circle passings through the origin and having intercepts a and b on real and imaginary axis, respectively, on the argand plane is `Re ((z-a)/(z-ib)) = 0`

Explanation

To show that the equation of a circle passing through the origin and having intercepts \( a \) and \( b \) on the real and imaginary axes respectively on the Argand plane is given by

\[ \text{Re} \left( \frac{z - a}{z - ib} \right) = 0, \]

we can follow these steps:

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