Question
Let complex numbers $$\alpha \,\,{\text{and }}\frac{1}{{\overline \alpha }}$$ lie on circles $${\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = {r^2}\,\,{\text{and}}\,\,\,{\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = 4{r^2}$$ respectively. If $${z_0} = {x_0} + i{y_0}$$ satisfies the equation $$2{\left| {{z_0}} \right|^2} = {r^2} + 2,{\text{then}}\,\left| \alpha \right| = $$
A.
$$\frac{1}{{\sqrt 2 }}$$
B.
$$\frac{1}{2}$$
C.
$$\frac{1}{{\sqrt 7 }}$$
D.
$$\frac{1}{3}$$
Answer :
$$\frac{1}{{\sqrt 7 }}$$
Solution :
As $$\alpha $$ lies on the circle $${\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = {r^2}$$
$$\therefore \,\,{\left| {\alpha - {z_0}} \right|^2} = {r^2}$$
$$\eqalign{
& \Rightarrow \,\,\left( {\alpha - {z_0}} \right)\left( {\overline a - {{\overline z }_0}} \right) = {r^2} \cr
& \Rightarrow \,\,\alpha \overline \alpha - \alpha {\overline z _0} - \overline \alpha {z_0} + {z_0}{\overline z _0} = {r^2} \cr
& \Rightarrow \,\,{\left| \alpha \right|^2} + {\left| {{z_0}} \right|^2} - \alpha {\overline z _0} - \overline \alpha {z_0} = {r^2}\,\,\,\,\,\,\,\,\,.....\left( {\text{i}} \right) \cr} $$
Also $$\frac{1}{{\overline \alpha }}$$ lies on the circle $${\left( {x - {x_0}} \right)^2} + {\left( {y - {y_0}} \right)^2} = 4{r^2}$$
$$\eqalign{
& \therefore \,\,{\left| {\frac{1}{{\overline \alpha }} - {z_0}} \right|^2} = 4{r^2} \cr
& \Rightarrow \,\,\left( {\frac{1}{{\overline \alpha }} - {z_0}} \right)\left( {\frac{1}{\alpha } - {{\overline z }_0}} \right) = 4{r^2} \cr
& \Rightarrow \,\,\frac{1}{{\alpha \overline \alpha }} - \frac{{{z_0}}}{\alpha } - \frac{{{{\overline z }_0}}}{{\overline \alpha }} + {z_0}{\overline z _0} = 4{r^2} \cr
& \Rightarrow \,\,\frac{1}{{{{\left| \alpha \right|}^2}}} - \frac{{{z_0}\overline \alpha }}{{{{\left| \alpha \right|}^2}}} - \frac{{{{\overline z }_0}\alpha }}{{{{\left| \alpha \right|}^2}}} + {\left| {{z_0}} \right|^2} = 4{r^2} \cr
& \Rightarrow \,\,1 + {\left| \alpha \right|^2}{\left| {{z_0}} \right|^2} - {z_0}\overline \alpha - {\overline z _0}\alpha = 4{r^2}{\left| \alpha \right|^2}\,\,\,\,\,\,\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Subtracting $${\text{e}}{{\text{q}}^n}$$ (i) from (i) we get
$$\eqalign{
& 1 - {\left| \alpha \right|^2} + {\left| {{z_0}} \right|^2}\left( {{{\left| \alpha \right|}^2} - 1} \right) = {r^2}\left( {4{{\left| \alpha \right|}^2} - 1} \right) \cr
& {\text{or }}\,\,\left( {{{\left| \alpha \right|}^2} - 1} \right)\left( {{{\left| {{z_0}} \right|}^2} - 1} \right) = {r^2}\left( {4{{\left| \alpha \right|}^2} - 1} \right) \cr
& {\text{Using }}{\left| {{z_0}} \right|^2} = \frac{{{r^2} + 2}}{2}{\text{ we get}} \cr
& \left( {{{\left| \alpha \right|}^2} - 1} \right)\frac{{{r^2}}}{r} = {r^2}\left( {4{{\left| \alpha \right|}^2} - 1} \right) \cr
& \Rightarrow \,\,{\left| \alpha \right|^2} - 1 = 8{\left| \alpha \right|^2} - 2 \cr
& \Rightarrow \,\,\left| \alpha \right| = \frac{1}{{\sqrt 7 }} \cr} $$