The circumcenter of the triangle formed by the lines $xy + 2x + 2y + 4 = 0$ and $x + y + 2 = 0$ is

Let $a$ be the length of a side of a square $OABC$ with $O$ being the origin. Its side $OA$ makes an acute angle $\alpha$ with the positive $x$-axis and the equations of its diagonals are $(\sqrt{3}+1)x+(\sqrt{3}-1)y=0$ and $(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0$. Then $a^2$ is equal to

The set of values that $\beta$ can assume so that the point $(0, \beta)$ lies on or inside the triangle formed by the lines $3x+y+2=0$,$2x-3y+5=0$,and $x+4y-14=0$ is

Let $A(h, k)$,$B(1, 1)$,and $C(2, 1)$ be the vertices of a right-angled triangle with $AC$ as the hypotenuse. If the area of the triangle is $1$,then which of the following can be the set of values for $k$?

In the figure given below,$ABCDEF$ is a regular hexagon of side length $1$,$AFPS$ and $ABQR$ are squares. Then,the ratio $\frac{\operatorname{ar}(APQ)}{\operatorname{ar}(SRP)}$ equals

Six consecutive sides of an equiangular octagon are $6, 9, 8, 7, 10, 5$ in that order. The integer nearest to the sum of the remaining two sides is