If $$\left| {z - 4} \right| < \left| {z - 2} \right|$$, its solution is given by :

The locus of the centre of a circle which touches the circle $$\left| {z - {z_1}} \right| = a$$ and$$\left| {z - {z_2}} \right| = b\,$$ externally

($$z,\,{z_1}\,\& \,{z_2}\,$$ are complex numbers) will be :

If $$\alpha \ne \beta $$ but $${\alpha ^2} = 5\alpha - 3$$ and $${\beta ^2} = 5\beta - 3$$ then the equation having $$\alpha /\beta $$ and $$\beta /\alpha \,\,$$ as its roots is

Product of real roots of equation $${t^2}{x^2} + \left| x \right| + 9 = 0$$