Let a given line $$L_1$$ intersects the x and y axes at P and Q, respectively. Let another line $$L_2$$, perpendicular to $$L_1$$, cut the x and y axes at R and S, respectively. Show that the locus of the point of intersection of the lines PS and QR is a circle passing through the origin.

Lines 5x + 12y - 10 = 0 and 5x - 12y - 40 = 0 touch a circle $$C_1$$ of diameter 6. If the centre of $$C_1$$ lies in the first quadrant, find the equation of the circle $$C_2$$ which is concentric with $$C_1$$ and cuts intercepts of length 8 on these lines.

The abscissa of the two points A and B are the roots of the equation $${x^2}\, + \,2ax\, - {b^2} = 0$$ and their ordinates are the roots of the equation $${x^2}\, + \,2px\, - {q^2} = 0$$. Find the equation and the radius of the circle with AB as diameter.

Through a fixed point (h, k) secants are drawn to the circle $$\,{x^2}\, + \,{y^2} = \,{r^2}$$. Show that the locus of the mid-points of the secants intercepted by the circle is $$\,{x^2}\, + \,{y^2} $$ = $$hx + ky$$.