The circumference of the circle $x^2 + y^2 - 2x + 8y - q = 0$ is bisected by the circle $x^2 + y^2 + 4x + 12y + p = 0$. Then $p + q$ is equal to

$P, Q$ and $R$ are the centres and $r_1, r_2, r_3$ are the radii respectively of three co-axial circles. Then $QRr_1^2 + RP r_2^2 + PQ r_3^2$ is equal to

If the straight line $x \cos \alpha + y \sin \alpha = P$ intersects the circle $x^2 + y^2 = a^2$ at $A$ and $B$,then the equation of the circle with diameter $\overline{AB}$ is

If $d$ is the distance between the centres of two circles,$r_1$ and $r_2$ are their radii,and $d = r_1 + r_2$,then

If the circles given by $S \equiv x^2+y^2-14x+6y+33=0$ and $S^{\prime} \equiv x^2+y^2-a^2=0$ $(a \in N)$ have $4$ common tangents,then the possible number of circles $S^{\prime}=0$ is

$A$ circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2+y^2=16$ and $x^2+y^2=1$. Then
$(A)$ radius of $S$ is $8$
$(B)$ radius of $S$ is $7$
$(C)$ centre of $S$ is $(-7,1)$
$(D)$ centre of $S$ is $(-8,1)$