Without changing the direction of the axes, the origin is shifted to the point $(2, 3)$. Then the equation $x^{2} + y^{2} - 4x - 6y + 9 = 0$ changes to

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 transformed equation of $3x^2 - 4xy = r^2$ when the coordinate axes are rotated through an angle $\tan^{-1}(2)$ is:

$A$ line $L$ has intercepts $a$ and $b$ on the coordinate axes. When the axes are rotated through a given angle $\theta$ keeping the origin fixed,this line $L$ has the intercepts $p$ and $q$. Then

If the origin is shifted to the point $(-2, 1)$,what are the new coordinates of the point $(4, -5)$?

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 $\alpha=$