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

The line $y=mx+c$ intercepts the circle $x^2+y^2=r^2$ in two distinct points,if

If $(6, -k)$ and $(-3, 2)$ are conjugate points with respect to the circle $x^2 + y^2 + 6x + 4y + 12 = 0$,then $k$ equals:

For the four circles $M, N, O$ and $P$,the following four equations are given:
Circle $M: x^2 + y^2 = 1$
Circle $N: x^2 + y^2 - 2x = 0$
Circle $O: x^2 + y^2 - 2x - 2y + 1 = 0$
Circle $P: x^2 + y^2 - 2y = 0$
If the centre of circle $M$ is joined with the centre of circle $N$,the centre of circle $N$ is joined with the centre of circle $O$,the centre of circle $O$ is joined with the centre of circle $P$,and lastly,the centre of circle $P$ is joined with the centre of circle $M$,then these lines form the sides of a:

The maximum area of a rectangle inscribed in the circle $(x+1)^{2}+(y-3)^{2}=64$ is

If a circle $S$ passing through the points $A(1, 2)$ and $B(2, 1)$ has its centre $C$ located in the third quadrant at a distance of $\frac{7}{\sqrt{2}}$ units from the line $AB$,then the point $P(1, -2)$: