Find the equation of the circle which passes through the origin and cuts off intercepts `3 a n d 4` from the positive parts of the axes respectively.

Explanation

In the given figure, we have
OA = 3, OB = 4 and C is the centre of the circle.
` therefore OL = (3)/(2) and CL = 2`
In `triangleOLC`, we have
`OC^(2) = OL^(2) + LC^(2)`
`rArr (OC)^(2) = ((3)/(2))^(2) + (2)^(2)`
`rArr OC = (5)/(2)`
Thus, the required circle has its centre at `((3)/(2),2)` and radius `(5)/(2)`.
Hence, its equation is `(x-(3)/(2))^(2) + (y-2)^(2) = ((5)/(2))^(2)`