The equation of the circle which cuts the circles $x^2+y^2+4x-7=0$,$2x^2+2y^2+3x+5y-9=0$,and $x^2+y^2+y=0$ orthogonally is

Statement $(A) :$ If two circles $x^2 + y^2 + 2gx + 2fy = 0$ and $x^2 + y^2 + 2g'x + 2f'y = 0$ touch each other,then $f'g = fg'$.
Reason $(R) :$ Two circles touch each other if the line joining their centers is perpendicular to all possible common tangents.

If the circle $x^2 + y^2 = 4$ bisects the circumference of the circle $x^2 + y^2 - 2x + 6y + a = 0$,then $a$ equals

Find the equation of a circle which passes through the point $(1,2)$ and the points of intersection of the circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2x-15=0$.

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