Find the $x$-intercept of the circle $x^2 + y^2 + 4x - 7y - 12 = 0$.

The angle between the circles $x^2+y^2+4x-14y+28=0$ and $x^2+y^2-12x-6y-4=0$ is

If $P$ is a point on the circle $x^{2}+y^{2}=4$, $Q$ is a point on the straight line $5x+y+2=0$ and $x-y+1=0$ is the perpendicular bisector of $PQ$, then $13$ times the sum of the abscissae of all such points $P$ is ........... .

If ${c^2} > {a^2}(1 + {m^2})$,then the line $y = mx + c$ will intersect the circle ${x^2} + {y^2} = {a^2}$ at:

Find the circumcenter of the triangle formed by the points $(a \cos \alpha, a \sin \alpha)$,$(a \cos \beta, a \sin \beta)$,and $(a \cos \gamma, a \sin \gamma)$.

If one end of a diameter of the circle $x^2 + y^2 - 3x + 8y - 4 = 0$ is $(6, -3)$,then what is the other end of the diameter?