Three distinct points A, B and C are given in the 2 -dimensional coordinates plane such that the ratio of the distance of any one of them from the point $$(1, 0)$$ to the distance from the point $$(-1, 0)$$ is equal to $${1 \over 3}$$. Then the circumcentre of the triangle ABC is at the point :

The lines $$p\left( {{p^2} + 1} \right)x - y + q = 0$$ and $$\left( {{p^2} + 1} \right){}^2x + \left( {{p^2} + 1} \right)y + 2q$$ $$=0$$ are perpendicular to a common line for :

The shortest distance between the line $$y - x = 1$$ and the curve $$x = {y^2}$$ is :

The remainder left out when $${8^{2n}} - {\left( {62} \right)^{2n + 1}}$$ is divided by 9 is :