If a straight line through $C(-\sqrt{8}, \sqrt{8})$ making an angle of $135^\circ$ with the $x$-axis cuts the circle $x = 5\cos\theta, y = 5\sin\theta$ at points $A$ and $B$,then the length of $AB$ is

The equation of the chord of the circle $x^2+y^2-4x=0$ whose midpoint is $(1,0)$ is

The ratio of the largest and shortest distances from the point $(2, -7)$ to the circle $x^2 + y^2 - 14x - 10y - 151 = 0$ is

$A$ line meets the coordinate axes at $A(a, 0)$ and $B(0, b)$. If the perpendicular distances from $A$ and $B$ to the tangent drawn at the origin to the circumcircle of $\triangle OAB$ are $m$ and $n$ respectively,then the diameter of that circle is

$A$ line $l$ meets the circle $x^2+y^2=61$ at points $A$ and $B$. Given that $P(-5, 6)$ is a point such that $PA=PB=10$,find the equation of line $l$.

Let the mirror image of a circle $c_{1}: x^{2}+y^{2}-2x-6y+\alpha=0$ in the line $y=x+1$ be $c_{2}: 5x^{2}+5y^{2}+10gx+10fy+38=0$. If $r$ is the radius of circle $c_{2}$,then $\alpha+6r^{2}$ is equal to: