If the two circles $${(x - 1)^2} + {(y - 3)^2} = {r^2}$$ and $${x^2} + {y^2} - 8x + 2y + 8 = 0$$ intersect in two distinct points, then

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 $$\left( {{x^2} + 4} \right){\left( {{{dy} \over {dx}}} \right)^2} = {n^2}\left( {{y^2} + 4} \right)$$