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 $$${{area\,\,of\,\,\Delta APQ} \over {area\,\,of\,\,\Delta ABC}} = {{\left( {...} \right)} \over {A{C^2}}}$$$

$$ABC$$ is a triangle with $$\angle B$$ greater than $$\angle C.\,D$$ and $$E$$ are points on $$BC$$ such that $$AD$$ is perpendicular to $$BC$$ and $$AE$$ is the bisector of angle $$A$$. Complete the relation $$$\angle DAE = {1 \over 2}\left[ {\left( {} \right) - \angle C} \right]$$$