The two circles ${x^2} + {y^2} - 4y = 0$ and ${x^2} + {y^2} - 8y = 0$

The equation of the circle having its centre on the line $x + 2y - 3 = 0$ and passing through the points of intersection of the circles ${x^2} + {y^2} - 2x - 4y + 1 = 0$ and ${x^2} + {y^2} - 4x - 2y + 4 = 0$, is

Two tangents are drawn from a point $P$ on radical axis to the two circles touching at $Q$ and $R$ respectively then triangle formed by joining $PQR$ is

$P$ is a point $(a, b)$ in the first quadrant. If the two circles which pass through $P$ and touch both the co-ordinate axes cut at right angles, then :

The number of common tangents to the circles ${x^2} + {y^2} = 4$ and ${x^2} + {y^2} - 6x - 8y = 24$ 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