If $x^2+y^2-4x-2y+5=0$ and $x^2+y^2-6x-4y-3=0$ are members of a coaxial system of circles,then the centre of a point circle in the system is

The equation of the radical axis of the circles $x^2 + y^2 + x - y + 2 = 0$ and $3x^2 + 3y^2 - 4x - 12 = 0$ is:

The radical centre of the circles $x^2+y^2-4x-6y+5=0$,$x^2+y^2-2x-4y-1=0$ and $x^2+y^2-6x-2y=0$ lies on the line

If the angle between the circles $x^2+y^2-4x-6y-3=0$ and $x^2+y^2+8x-4y+\lambda=0$ is $60^{\circ}$,then a value of $\lambda$ is

Suppose $A(2,3)$ and $B$ are the points of intersection of two circles. The points $P$ lying on one circle and $Q$ lying on the other circle are such that $BP$ and $BQ$ constitute the diameters of the circles. If the slopes of the radical axis and $PQ$ are $3/4$ and $a/b$ respectively,then the value of $3a+4b$ is

If the circles $x^{2}+y^{2}=9$ and $x^{2}+y^{2}+2 \alpha x+2 y+1=0$ touch each other internally,then $\alpha$ is equal to