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

If two circles which pass through the points $(0, a)$ and $(0, -a)$ and touch the line $y = mx + c$ cut orthogonally, then:

The equation of the circle passing through the points of intersection of the circles $x^2+y^2-2px=0$ and $x^2+y^2-2qy=0$ and having its centre on the line $\frac{x}{p}-\frac{y}{q}=2$ is:

The value of $\lambda$,for which the circle $x^2 + y^2 + 2\lambda x + 6y + 1 = 0$ intersects the circle $x^2 + y^2 + 4x + 2y = 0$ orthogonally is

The equations of three circles are $x^2 + y^2 - 12x - 16y + 64 = 0$,$3x^2 + 3y^2 - 36x + 81 = 0$,and $x^2 + y^2 - 16x + 81 = 0$. The coordinates of the point from which the lengths of the tangents drawn to each of the three circles are equal is:

The number of common tangents to the circles ${x^2} + {y^2} - 4x - 6y - 12 = 0$ and ${x^2} + {y^2} + 6x + 18y + 26 = 0$ is