If the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 10x + \lambda = 0$ touch externally,then $\lambda$ is equal to:

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 $\frac{\beta}{\alpha}=$

If $\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S \equiv x^2+y^2+\alpha x+6y=0$,$S^{\prime} \equiv x^2+y^2+2\alpha x+\alpha y+6=0$ and $S^{\prime\prime} \equiv x^2+y^2+6\alpha x-\alpha y+3=0$,then the distance between the radical centre and the centre of the circle $S^{\prime}=0$ is:

If the circle $S \equiv x^2+y^2+2gx+4y+1=0$ bisects the circumference of the circle $x^2+y^2-2x-3=0$,then the radius of circle $S=0$ is

The equation of the radical axis of the pair of circles $7x^2+7y^2-7x+14y+18=0$ and $4x^2+4y^2-7x+8y+20=0$ is

$(a, 0)$ and $(b, 0)$ are centers of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is: