$x^2+y^2+2x-6y-6=0$ and $x^2+y^2-6x-2y+k=0$ are two intersecting circles and $k$ is not an integer. If $\theta$ is the angle between the two circles and $\cos \theta = \frac{-5}{24}$,then $k=$

$A$ triangle $PQR$ is inscribed in the circle $x^2 + y^2 = 25$. If the coordinates of $Q$ and $R$ are $(3, 4)$ and $(-4, 3)$ respectively,then $\angle QPR = \dots$

If two circles $(x - 1)^2 + (y - 3)^2 = r^2$ and $x^2 + y^2 - 8x + 2y + 8 = 0$ intersect in two distinct points,then

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:

If the lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ are tangents to a circle,then the radius of the circle is

If the lengths of the chords intercepted by the circle $x^2 + y^2 + 2gx + 2fy = 0$ from the coordinate axes are $10$ and $24$ respectively,then the radius of the circle is: