If a circle $S$ passing through the origin and having its centre on the line $x-y=0$ cuts the circle $x^2+y^2-4x-6y+10=0$ orthogonally,then the diameter of $S$ is

Let the equation of the circle,which touches the $x$-axis at the point $(a, 0), a > 0$ and cuts off an intercept of length $b$ on the $y$-axis,be $x^2 + y^2 - \alpha x + \beta y + \gamma = 0$. If the circle lies below the $x$-axis,then the ordered pair $(2a, b^2)$ is equal to:

The area (in sq. units) of the region bounded by the circle $x^2+y^2=64$, positive $x$-axis and the line $y=\sqrt{3}x$ is

Let $ABC$ be a triangle with $AB=1$,$AC=3$,and $\angle BAC=\frac{\pi}{2}$. If a circle of radius $r>0$ touches the sides $AB$,$AC$ and also touches the circumcircle of triangle $ABC$ internally,then the value of $r$ is:

If one of the diameters of the circle given by the equation $x^{2}+y^{2}+4x+6y-12=0$ is a chord of a circle $S$ whose centre is $(2,-3)$, then the radius of $S$ is:

If the line $x + 2by + 7 = 0$ is a diameter of the circle $x^2 + y^2 - 6x + 2y = 0$,then $b = $