If the centre $(\alpha, \beta)$ of a circle cutting the circles $x^2+y^2-2y-3=0$ and $x^2+y^2+4x+3=0$ orthogonally lies on the line $2x-3y+4=0$,then $2\alpha+\beta=$

The equation of the radical axis of the circles $x^2+y^2+4x+6y+7=0$ and $4x^2+4y^2+8x+12y-9=0$ is:

The equation of the circle passing through the origin and cutting the circles $x^2+y^2+6x-15=0$ and $x^2+y^2-8y-10=0$ orthogonally is

If a circle passing through the point $(1,1)$ cuts the circles $x^2+y^2+4x-5=0$ and $x^2+y^2-4y+3=0$ orthogonally,then the center of that circle is:

If $x^2+y^2-a^2+\lambda(x \cos \alpha+y \sin \alpha-p)=0$ is the smallest circle through the points of intersection of $x^2+y^2=a^2$ and $x \cos \alpha+y \sin \alpha=p$,where $0 < p < a$,then $\lambda=$

The circle ${x^2 + y^2 + 2gx + 2fy + c = 0}$ bisects the circumference of the circle ${x^2 + y^2 + 2g'x + 2f'y + c' = 0}$,if