Let A (0,0) B (4,0) and C (0,6)
AB has a constant ordinate and
hence t is on x axis (ordinate=0)
AC has a constant abscissa (=0)
Hence it is on y axis
⇒AB⊥AC
Hence the given triangle is a
right angled triangle ,at ∠A
so, BC is the hypotenuse,
and circumferential is the midpoint
of the hypotenuse.
Hence . midpoint of BC
(4+02,0+62)
(42,62)=(2,3)
∴ The circumcentre is (2,3)