The equation of a line making an angle $60^{\circ}$ with the line $x+y-3=0$ and passing through the point $(1,1)$ is
Let $P$ be the pair of lines represented by $2 x^2-5 x y+2 y^2+6 x-3 y=0$ and consider the following independent statements
(i) $\alpha$ is the $x$ coordinate of the point of intersection of the pair of lines $P$.
(ii) $\beta$ is the slope of one of the lines of $P$ passing through origin.
(iii) $\gamma$ is the constant term in the equation of the pair of angular bisectors of $P$.
Then,
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
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