When the coordinate axes are rotated about the origin in the positive direction through an angle $\frac{\pi }{4}$, if the equation $49x^2+25y^2=1225$ is transformed to $p{x}^2+qxy+r{y}^2=t$ and the G.C.D of $p,q,r,t$ is $1$, then

If ${\theta }_1,{\theta }_2,{\theta }_3$ are respectively the angles by which the coordinate axes are to be rotated to eliminate the $xy$ term from the following equations, then the descending order of these angles is

$A_1=3x^2+5xy+3y^2+2x+3y+4=0$
$A_2=5x^2+2\sqrt{3}xy+3y^2+6=0$

$A_3=4x^2+\sqrt{3}xy+5y^2-4=0$

If the equation $x^2+y^2-4x-6y-12=0$ is transformed to $X^2+Y^2=25$ when the axes are translated to a point then the new coordinates of (-3, 5) are

The angle of rotation of the axes so that the equation $x+y-6=0$ may be reduced to the form $X=3\sqrt{2}$ is

The new coordinates of a point $\left(4,5\right)$, when the origin is shifted to the point $\left(1,-2\right)$ are

Without changing the direction of coordinate axes, origin is transferred to (h, k), so that the linear (one degree) terms in the equation $x^2+y^2-4x+6y-7=0$ are eliminated. Then the point (h, k) is