$A$ 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 equation $8x^2 + 8xy + 2y^2 + 26x + 13y + 15 = 0$ represents a pair of parallel straight lines. The distance between them is

The distance between the parallel lines given by $(x+7y)^2 + 4\sqrt{2}(x+7y) - 42 = 0$ 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 $k=$

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