If the circle $x^2 + y^2 = 4$ bisects the circumference of the circle $x^2 + y^2 - 2x + 6y + a = 0$,then $a$ equals

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 ............. $sq. \, units$.

The equation of the line joining the centres of the circles belonging to the coaxial system of circles $4x^2 + 4y^2 - 12x + 6y - 3 + \lambda(x + 2y - 6) = 0$ is

If the circles $(x-2)^2+(y-3)^2=25$ and $25x^2+25y^2-40x-70y-160=0$ touch internally at $(\alpha, \beta)$,then $\alpha+\beta=$

Find the equation of a circle with radius $5$ units and touching the circle $x^2+y^2-2x-4y-20=0$ at the point $(5,5)$.

$A$ circle $S$ cuts three circles $x^2+y^2-4x-2y+4=0$,$x^2+y^2-2x-4y+1=0$,and $x^2+y^2+4x+2y+1=0$ orthogonally. Then,the radius of $S$ is