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 :

If $$P$$ and $$Q$$ are the points of intersection of the circles
$${x^2} + {y^2} + 3x + 7y + 2p - 5 = 0$$ and $${x^2} + {y^2} + 2x + 2y - {p^2} = 0$$ then there is a circle passing through $$P,Q $$ and $$(1, 1)$$ for :

The point diametrically opposite to the point $$P(1, 0)$$ on the circle $${x^2} + {y^2} + 2x + 4y - 3 = 0$$ is :

The differential equation of the family of circles with fixed radius $$5$$ units and centre on the line $$y = 2$$ is :