The two circles $x^2 + y^2 - 2x + 6y + 6 = 0$ and $x^2 + y^2 - 5x + 6y + 15 = 0$:

The points of intersection of the line $ax + by = 0$ $(a \neq b)$ and the circle $x^2 + y^2 - 2x = 0$ are $A(\alpha, 0)$ and $B(1, \beta)$. The image of the circle with $AB$ as a diameter in the line $x + y + 2 = 0$ is:

Find the equation of the chord of the circle $x^2 + y^2 - 6x + 8y = 0$ which is bisected at the point $(5, -3)$.

If the angle between the circles $x^2+y^2-12x-6y+41=0$ and $x^2+y^2+kx+6y-59=0$ is $45^{\circ}$,then a value of $k$ is

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?

If $(3,-2)$ is the centre of the circle $S \equiv x^2+y^2+2gx+2fy-23=0$ and $A$ is a point on the circle $S=0$ such that its distance from a point $P(-1,-5)$ is least,then $A=$