When the origin is shifted to $(2, 3)$,the transformed equation of a curve becomes $x^2+3xy-2y^2+17x-7y-11=0$. Find the original equation of the curve.

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

The point $P(\alpha, \beta)$ with $\alpha > 0, \beta > 0$ undergoes the following transformations successively:
$a)$ Translation by $3$ units in the positive direction of the $x$-axis.
$b)$ Reflection about the line $y = -x$.
$c)$ Rotation of axes through an angle of $\frac{\pi}{4}$ about the origin in the positive direction.
If the final position of the point $P$ is $(-4\sqrt{2}, -2\sqrt{2})$,then find the value of $(\alpha + \beta)$.

If the axes are rotated by an angle of $30^{\circ}$ in the negative direction (clockwise) while keeping the origin fixed,what are the new coordinates of the point $(2, 1)$?

To which point should the origin be shifted,without changing the direction of the axes,so that the equation $x^2 + y^2 - 4x + 6y - 7 = 0$ transforms into an equation that contains no first-degree terms?

By rotating the coordinate axes in the positive direction about the origin by an angle $\alpha$,if the point $(1,2)$ is transformed to $\left(\frac{3 \sqrt{3}-1}{2 \sqrt{2}}, \frac{\sqrt{3}+3}{2 \sqrt{2}}\right)$ in the new coordinate system,then $\alpha=$