The equation of the circle which touches the $$x$$-axis, passes through the point $$(1,1)$$ and whose centre lies on the line $$x+y=3$$ in the first quadrant is

The equation of a circle passing through the origin is $$x^2+y^2-6 x+2 y=0$$. The equation of one of its diameter is

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