If all chords of the curve $2 x^2-y^2+3 x+2 y=0$, which subtend a right angle at the origin always passing through the point $(\alpha, \beta)$, then $(\alpha, \beta)=$

If the origin is shifted to remove the first degree terms from the equation $2 x^2-3 y^2+4 x y+4 x+4 y-14=0$, then with respect to this new coordinate system the transformed equation of $x^2+y^2-3 x y+4 y+3=0$ is

The circumcentre of the triangle formed by the lines $x+y+2=0,2 x+y+8=0$ and $x-y-2=0$ is

If the line $2 x-3 y+5=0$ is the perpendicular bisector of the line segment joining $(1,-2)$ and $(\alpha, \beta)$, then $\alpha+\beta=$