If $(\alpha, \beta)$ is the external centre of similitude of the circles $x^2+y^2=3$ and $x^2+y^2-2x+4y+4=0$,then $\frac{\beta}{\alpha}=$

Consider the system of circles $x^2+y^2+2fy+\lambda(x^2+y^2+2gx+k)=0$,where $g \neq 0, f \neq 0$ and $\lambda$ is a parameter. If $A$ and $B$ are the point circles of this system such that $\angle AOB = \frac{\pi}{2}$,then $g^2$ is equal to

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:

If one of the two circles $x^2+y^2+\alpha_1(x-y)+c=0$ and $x^2+y^2+\alpha_2(x-y)+c=0$ lies within the other,then (where $\alpha_1, \alpha_2 \in R, \alpha_1 \neq \alpha_2$):

If the length of the transverse common tangent to the circles $x^{2} + y^{2} = 1$ and $(x - h)^{2} + y^{2} = 1$ is $2\sqrt{3}$,find the value of $h$.

If the circles $x^2+y^2-6x-8y-12=0$ and $x^2+y^2-4x+6y+k=0$ are orthogonal to each other,then the value of $k$ is: