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

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 $(-8,1)$

If the circles ${x^2}\, + {y^2}\, - 16x\, - 20y\, + \,164\,\, = \,\,{r^2}$ and ${(x - 4)^2} + {(y - 7)^2} = 36$ intersect at two distinct points, then

Let the equation $x^{2}+y^{2}+p x+(1-p) y+5=0$ represent circles of varying radius $\mathrm{r} \in(0,5]$. Then the number of elements in the set $S=\left\{q: q=p^{2}\right.$ and $\mathrm{q}$ is an integer $\}$ is ..... .

The number of common tangents of the circles given by $x^2 +y^2 - 8x - 2y + 1 = 0$ and $x^2 + y^2 + 6x + 8y = 0$ is

The circles ${x^2} + {y^2} + 4x + 6y + 3 = 0$ and $2({x^2} + {y^2}) + 6x + 4y + C = 0$ will cut orthogonally, if $C$ equals