Consider a family of circles passing through two fixed points A (3, 7) and B (6, 5). Show that the chords on which the circle $${x^2}\, + \,{y^2} - \,4x - \,6y - 3 = 0$$ cuts the members of the family are concurrent at a point. Find the coordinate of this point.

Tagent at a point $${P_1}$$ {other than $$(0, 0)$$} on the curve $$y = {x^3}$$ meets the curve again at $${P_2}$$. The tangent at $${P_2}$$ meets the curve at $${P_3}$$, and so on. Show that the abscissae of $${P_1},\,{P_2},{P_3}......{P_n},$$ form a G.P. Also find the ratio.

[area $$\left( {\Delta {P_1},{P_2},{P_3}} \right)$$]/[area $$\left( {{P_2},{P_3},{P_4}} \right)$$]

Find the coordinates of the point at which the circles $${x^2}\, + \,{y^2} - \,4x - \,2y = - 4\,\,and\,\,{x^2}\, + \,{y^2} - \,12x - \,8y = - 36$$ touch each other. Also find equations common tangests touching the circles in the distinct points.

A line through $$A (-5, -4)$$ meets the line $$x + 3y + 2 = 0,$$ $$2x + y + 4 = 0$$ and $$x - y - 5 = 0$$ at the points $$B, C$$ and $$D$$ respectively. If $${\left( {15/AB} \right)^2} + {\left( {10/AC} \right)^2} = {\left( {6/AD} \right)^2},$$ find the equation of the line.