The two circles $x^2 + y^2 - 2x - 3 = 0$ and $x^2 + y^2 - 4x - 6y - 8 = 0$ are such that:

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 coordinates of the point from where the tangents drawn to the circles ${x^2} + {y^2} = 1$,${x^2} + {y^2} + 8x + 15 = 0$,and ${x^2} + {y^2} + 10y + 24 = 0$ are of the same length are:

Two tangents are drawn from a point $P$ on the radical axis to two circles,touching them at $Q$ and $R$ respectively. Then,the triangle formed by joining $P, Q,$ and $R$ is:

If $(p, q)$ is the centre of the circle which cuts the three circles $x^2+y^2-2x-4y+4=0$,$x^2+y^2+2x-4y+1=0$ and $x^2+y^2-4x-2y-11=0$ orthogonally,then $p+q=$

The radical axis of the circles $x^2+y^2+2gx+2fy+c=0$ and $2x^2+2y^2+3x+8y+2c=0$ touches the circle $x^2+y^2+2x+2y+1=0$. Then