Question

Find the centre and radius of the circle formed by all the points represented by $z=x+iy$ satisfying the relation $\frac{|z-\alpha |}{|z-\beta |}=k(k\ne 1)$ where $\alpha$ and $\beta$ are constant complex numbers given by $\alpha ={\alpha }_1+{i\alpha }_2$, $\beta ={\beta }_1+{i\beta }_2$.

Answer 1

Squaring the given relation, we have
$(z-\alpha )\left(\overline{z-\alpha }\right)=k^2(z-\beta )\left(\overline{z-\beta }\right)$
or $(z-\alpha )(\overline{z}-\overline{\alpha })=k^2(z-\beta )(\overline{z}-\overline{\beta })$
or $z\overline{z}-\overline{\alpha }z-\alpha \overline{z}+\alpha \overline{\alpha }=k^2(z\overline{z}-\overline{\beta }z-\beta \overline{z}+\beta \overline{\beta })$
or $z\overline{z}(1-k^2)-(\overline{\alpha }-k^2\overline{\beta })z-(\alpha -k^2\beta )\overline{z}+\alpha \overline{\alpha }-k^2\beta \overline{\beta }=0$
or $z\overline{z}-\frac{\overline{\alpha }-k^2\overline{\beta }}{1-k^2}z-\frac{(\alpha -k^2\beta )}{1-k^2}\overline{z}+\frac{{\left|\alpha \right|}^2-k^2{\left|\beta \right|}^2}{1-k^2}=0$
or $z\overline{z}-\overline{a}z-a\overline{z}+a\overline{a}-a\overline{a}+\frac{{\left|\alpha \right|}^2-k^2{\left|\beta \right|}^2}{1-k^2}=0$
or ${|z-a|}^2={\left|a\right|}^2-\frac{{\left|\alpha \right|}^2-k^2\left|{\beta }^2\right|}{1-k^2}=r^2$
Above represents a circle with center at $a=\frac{\alpha -k^2\beta }{1-k^2}$ and $r^2=$ as written above.