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

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

Consider a family of circles which are passing through the point $$(-1, 1)$$ and are tangent to $$x$$-axis. If $$(h, k)$$ are the coordinate of the centre of the circles, then the set of values of $$k$$ is given by the interval :