The area (in sq units) of the minor segment bounded by the circle $$x^2+y^2=a^2$$ and the line $$x=\frac{a}{\sqrt{2}}$$ is

The points of intersection of circles $$(x+1)^2+y^2=4$$ and $$(x-1)^2+y^2=9$$ are $$(a, \pm b)$$, then $$(a, b)$$ equals to

The circle $$x^2+y^2+3 x-y+2=0$$ cuts an intercept on $$X$$-axis of length

$$S \equiv x^2+y^2-2 x-4 y-4=0$$ and $$S^{\prime} \equiv x^2+y^2-4 x-2 y-16=0$$ are two circles the point $$(-2,-1)$$ lies