Question

Find the equation of the circle which passes through the origin and cuts off chords of lengths 4 and 6 on the positive side of the x-axis and y-axis respectively.

Answer 1

We have a circle that passes through origin O(0, 0) and cut off on intersept of length 4 units on x-axis and 6 units on y-axis.
That is, OA = 4
OB = 6
C - be the centre of the circle and CM and CN are perpendicular line drawn on OA and OB respectively.
Coordinates of A am (4, 0) and B = (0, 6)
$\therefore$ Coordinates of M = (2, 0) and N = (0, 3)
Thus coordinates of C = (2, 3)
Now in $\Delta OCM$
$O{C}^2=O{M}^2+C{M}^2=2^2+3^2[\because CM=ON=3]=4+9\therefore OC=\sqrt{13}$
Thus, the required circle is
$(x-2{)}^2+(y-3{)}^2=13x^2+y^2-4x-6y=0$