The number of common tangents to the circles $x^2+y^2+4x-6y-12=0$ and $x^2+y^2-8x+10y+5=0$ is

$A$ tangent $PT$ is drawn to the circle $x^2 + y^2 = 4$ at the point $P(\sqrt{3}, 1)$. $A$ line $L$ perpendicular to $PT$ is a tangent to the circle $(x - 3)^2 + y^2 = 1$. The possible equation of $L$ is:

If the radical centre of the three circles $x^2+y^2=1$,$x^2+y^2-2x-3=0$,and $x^2+y^2-2y-3=0$ is $C(\alpha, \beta)$ and $r$ is the sum of the radii of the given circles,then the equation of the circle with $C(\alpha, \beta)$ as centre and $r$ as radius is:

The point $(3, -4)$ lies on both the circles $x^2 + y^2 - 2x + 8y + 13 = 0$ and $x^2 + y^2 - 4x + 6y + 11 = 0$. Then,the angle between the circles is

The radical centre of the three circles $x^2+y^2-1=0$,$x^2+y^2-8x+15=0$ and $x^2+y^2+10y+24=0$ is

The angle between circles $x^2+y^2+2x+4y+1=0$ and $x^2+y^2-2x+6y-3=0$ is