A triangle has two of its sides along the axes, its third side touches the circle x2+ y2- 2ax - 2ay + a2= 0Prove that th

Solution

Let OPQ be a triangle whose two sides OP and OQ are along the coordinate axes and the third side PQ touches the circle

x2 + y2 − 2ax − 2ay + a2 = 0 …(i)

whose centre is (a, a) and radius =

Since the ΔPOQ is right angled at O, therefore the circle drawn on PQ as diameter will pass through O and so the circumcentre of the ΔPOQ is the middle point R of PQ. Let (h, k) be the coordinates of the point R whose locus is to be found, Then

…(iii)

Since the straight lien (ii) i.e., the line qx + py − pq = 0 touches the circle (i), therefore the length of the perpendicular drawn to this line from the centre (a, a) of the circle (i) is equal to the radius a of the circle (i).

∴    or [a (p + q) − pq]2 = a2 (p2 + q2)

or (p + q)2 + p2q2 − 2pqa (p + q) = a2 (p2 + q2)

or 2a2pq + p2q2 − 2pqa (p + q) = 0

or 2a2 − 2a (p + q) + pq = 0, dividing by pq

or 2a2 − 2a (2h + 2k) + 2h . 2k = 0, from (iii).

∴  locus of the centre R(h, k) of the circumcircle of ΔOPQ is

a2 − 2a (x + y) + 2xy = 0.