$A$ 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

Let $3x^2+8xy-3y^2=0$ represent the lines $L_1, L_2$ and $3x^2+8xy-3y^2+2x-4y-1=0$ represent the lines $L_3, L_4$. Let $L$ be the line joining the points of intersection of $L_1, L_3$ and $L_2, L_4$. Then,the area (in sq units) of the triangle formed by $L$ with the coordinate axes is

The acute angle between the pair of straight lines joining the origin to the points of intersection of the line $x+y-1=0$ with the pair of straight lines $k x^2+8 x y-3 y^2+2 x-4 y-1=0$ 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 $2c^2-15c =$

The line $x+y=k$ meets the pair of straight lines $x^2+y^2-2x-4y+2=0$ at two points $A$ and $B$. If $O$ is the origin and $\angle AOB=90^{\circ}$,then the value of $k (>1)$ is

The angle between the pair of straight lines formed by joining the points of intersection of $x^2+y^2=4$ and $y=3x+c$ to the origin is a right angle. Then $c^2$ is equal to