The combined equation of the diagonals of the parallelogram formed by the lines

$$ \left(7 x^2-4 x y+8 y^2\right)^2+(4 x-8 y-32)\left(7 x^2-4 x y+8 y^2\right)=0 $$

is

If the origin lies on a diameter of the circle $x^2+y^2-4 x-2 y-4=0$, then the equation of the circle passing through the end points of that diameter and the point $(1,2)$ is

If $\alpha \neq-4$ and $(2, \alpha)$ is the mid-point of a chord of the circle $x^2+y^2-4 x+8 y+6=0$, then the values of the $y$-intercept of the chord lie in the interval

$C_1$ and $C_2$ are the external and internal centres of similitude of the circles $x^2+y^2-2 x+4 y+1=0$ and $x^2+y^2+4 x-6 y+12=0$. If the radius of the circle having $C_1 C_2$ as its diameters is $r$, then $\frac{9}{2} r=$