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

The radical centre of the circles ${x^2} + {y^2} - 16x + 60 = 0,\,{x^2} + {y^2} - 12x + 27 = 0,$ ${x^2} + {y^2} - 12y + 8 = 0$ is

Let $C_1$ and $C_2$ be the centres of the circles $x^2 + y^2 -2x -2y -2 = 0$ and $x^2 + y^2 - 6x-6y + 14 = 0$ respectively. If $P$ and $Q$ are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $PC_1QC_2$ is ............. $\mathrm{sq. \, units}$

The centre of the circle passing through $(0, 0)$ and $(1, 0)$ and touching the circle ${x^2} + {y^2} = 9$ is

Radius of circle touching $y-$axis at point $P(0,2)$ and circle $x^2 + y^2 = 16$ internally-

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