The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle of area 154 sq. units. Then the equation of this circle is

The line x + 3y = 0 is a diameter of the circle $${x^2} + {y^2} - 6x + 2y = 0\,$$.

If $$x = \sec \theta - \cos \theta $$ and $$y = {\sec ^n}\theta - {\cos ^n}\theta $$, then show
that $$\left( {{x^2} + 4} \right){\left( {{{dy} \over {dx}}} \right)^2} = {n^2}\left( {{y^2} + 4} \right)$$

The greater of the two angles $$A = 2{\tan ^{ - 1}}\left( {2\sqrt 2 - 1} \right)$$ and $$B = 3{\sin ^{ - 1}}\left( {1/3} \right) + {\sin ^{ - 1}}\left( {3/5} \right)$$ is ________ .