A square is inscribed in the circle $${x^2} + {y^2} - 2x + 4y + 3 = 0$$. Its sides are parallel to the coordinate axes. The one vertex of the square is

Two circles $${x^2} + {y^2} = 6$$ and $${x^2} + {y^2} - 6x + 8 = 0$$ are given. Then the equation of the circle through their points of intersection and the point (1, 1) is

Given $$y = {{5x} \over {3\sqrt {{{\left( {1 - x} \right)}^2}} }} + {\cos ^2}\left( {2x + 1} \right)$$; Find $${{dy} \over {dx}}$$.

$$ABC$$ is a triangle, $$P$$ is a point on $$AB$$, and $$Q$$ is point on $$AC$$ such that $$\angle AQP = \angle ABC$$. Complete the relation $$${{area\,\,of\,\,\Delta APQ} \over {area\,\,of\,\,\Delta ABC}} = {{\left( {...} \right)} \over {A{C^2}}}$$$