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 $2c^2-15c =$

The area (in sq. units) of the triangle formed by the lines $x^2-3xy+y^2=0$ and $x+y+1=0$ is

If the pair of lines joining the origin and the points of intersection of the line $ax+by=1$ and the curve $x^2+y^2-x-y-1=0$ are at right angles,then the locus of the point $(a, b)$ is a circle of radius

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 two lines represented by the pair of straight lines $9x^2 - 24xy + 16y^2 + 3x - 4y - 6 = 0$ is:

The line $x+2y=k$ meets the curve $2x^2-2xy+3y^2+2x-y-1=0$ at two points $A$ and $B$. Let $O$ be the origin. If the line segments $OA$ and $OB$ are perpendicular to each other,then $k=$