The number of common tangents to two circles ${x^2} + {y^2} = 4$ and ${x^2} - {y^2} - 8x + 12 = 0$ is

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 ..... .

Two circles ${S_1} = {x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$ and ${S_2} = {x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$ cut each other orthogonally, then

The equation of the circle which intersects circles ${x^2} + {y^2} + x + 2y + 3 = 0$, ${x^2} + {y^2} + 2x + 4y + 5 = 0$and ${x^2} + {y^2} - 7x - 8y - 9 = 0$ at right angle, will be

The equation of the circle passing through the point $(-2, 4)$ and through the points of intersection of the circle ${x^2} + {y^2} - 2x - 6y + 6 = 0$ and the line $3x + 2y - 5 = 0$, is

The number of common tangents, to the circles $x^2+y^2-18 x-15 y+131=0$ and $x^2+y^2-6 x-6 y-7=0$, is :