Question

A straight rod AB of fixed length c moves so that its extremities A , B lie on two fixed mutually perpendicular lines . Prove that the locus of the circumcenter of triangle OAB is a circle , where O is the the point of intersection of mutually perpendicular straight lines

Answer 1

Let A (a, 0), B (0, b) be the points on the axes then $\angle$AOB = $\pi$/2 therefore circle OAB is on AB as diameter whose equation is
(x-a) (x-0) + (y-0) (y-b) = 0
Or $x^2+y^2-ax-by=0$
If (h, k) be the centre then h = a/2, k = b/2.
Since AB = c $\therefore a^2+b^2=c^2$
or 4h${}^2+4k^2=c^2$
$\therefore$ Locus of the centre is $x^2+y^2=c^2/4$