The equation of the common chord of the circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2-5x-6y+4=0$ is

Two circles whose radii are equal to $4$ and $8$ intersect at right angles. The length of their common chord is:

The equation of the transverse common tangent of the circles $x^2+y^2-6x-8y+9=0$ and $x^2+y^2+2x-2y+1=0$ is

If the equation of the circle having the common chord of the circles $x^2+y^2+x-3y-10=0$ and $x^2+y^2+2x-y-20=0$ as its diameter is $x^2+y^2+\alpha x+\beta y+\gamma=0$,then $\alpha+2\beta+\gamma=$

If a variable circle $S=0$ touches the line $y=x$ and passes through the point $(0,0)$,then the fixed point that lies on the common chord of the circles $x^2+y^2+6x+8y-7=0$ and $S=0$ is

If the chord of contact of tangents from a point $A$ to a given circle passes through $B$,then the circle with $AB$ as a diameter will . . . . . .