The distance between the pair of lines $x^2+2 \sqrt{2} xy+2y^2+4x+4 \sqrt{2}y+1=0$ is

The line $x+2y-c=0$ meets the curve $x^2+y^2-3x-6y+3=0$ at two points $P$ and $Q$ and $\angle POQ = \frac{\pi}{2}$,where $O$ is the origin. Then $2c^2-15c =$

Consider the lines $L_1 \equiv 4x + 5y - 6 = 0$,$L_2 \equiv 2x + 3y - 4 = 0$,and $L_3 \equiv 3x - y + 2 = 0$. If the line $L_1 = 0$ intersects the lines $L_2 = 0$ and $L_3 = 0$ at the points $A$ and $B$ respectively,then the combined equation of lines $OA$ and $OB$ is

The distance between the lines represented by the equation $x^2 + 2\sqrt{3}xy + 3y^2 - 3x - 3\sqrt{3}y - 4 = 0$ is

The combined equation of the bisectors of the angles between the lines joining the origin to the points of intersection of the curve $x^2+y^2+xy+x+3y+1=0$ and the line $x+y+2=0$ is

$A$ pair of perpendicular lines passes through the origin and also through the points of intersection of the curve $x^2+y^2=4$ with $x+y=a$,where $a>0$. Then $a$ is equal to