Question Number 19555 by Tinkutara last updated on 12/Aug/17
$$\mathrm{Find}\:\mathrm{center}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{having} \\ $$$$\mathrm{equation}\:{z}\bar {{z}}\:+\:\left(\mathrm{1}\:−\:{i}\right){z}\:+\:\left(\mathrm{1}\:+\:{i}\right)\bar {{z}}\:−\:\mathrm{1}\:=\:\mathrm{0}. \\ $$
Answered by ajfour last updated on 12/Aug/17
$$\mathrm{Equation}\:\mathrm{of}\:\mathrm{circle}: \\ $$$$\mid\mathrm{z}−\mathrm{z}_{\mathrm{0}} \mid=\mathrm{r} \\ $$$$\Rightarrow\:\:\:\left(\mathrm{z}−\mathrm{z}_{\mathrm{0}} \right)\left(\bar {\mathrm{z}}−\bar {\mathrm{z}}_{\mathrm{0}} \right)=\mathrm{r}^{\mathrm{2}} \\ $$$$\mathrm{or}\:\:\:\:\mathrm{z}\bar {\mathrm{z}}−\bar {\mathrm{z}}_{\mathrm{0}} \mathrm{z}−\mathrm{z}_{\mathrm{0}} \bar {\mathrm{z}}+\mid\mathrm{z}_{\mathrm{0}} \mid^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{comoparing}\:\mathrm{this}\:\mathrm{with}\:\mathrm{given}\:\mathrm{eqn}. \\ $$$$\:\:\:\:\mathrm{z}_{\mathrm{0}} =\mathrm{1}+\mathrm{i}\:\:\:\mathrm{and}\:\:\mathrm{r}=\sqrt{\mathrm{1}+\mid\mathrm{z}_{\mathrm{0}} \mid^{\mathrm{2}} }\:=\sqrt{\mathrm{3}}\:. \\ $$
Commented by Tinkutara last updated on 12/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$