The locus of the centre of the circle passing through the origin O and the points of intersection A and B of any line through (a, b) and the coordinate axes is

Options

A. `(x)/(a)+(y)/(b)=1`
B. `(a)/(x)+(b)/(y)=1`
C. `(x)/(a)+(y)/(b)=2`
D. `(a)/(x)+(b)/(y)=2`

Answer

D

Explanation

Let the coordinates of A and B be (p, 0) and (0, q) respectively. Then, equation of AB is
`(x)/(p)+(y)/(q)=1`
Since it passes through (a, b).
`:. (a)/(p)+(b)/(q)=1" " ..(i)`
The triangle OAB is a right-angled triangle . So, it is a diameter of the circle a passing through O, A and B. So , coordinates of the centre of the circle are (p/2, q/2).
Let (h, k) be the centre of the circle. Then,
`h=p//2, k=q//2 rArr p=2h, q=2k`
Substituting values of p, q in (i), we get:
`(a)/(2h)+(b)/(2k)=1`
Hence, the locus of (h, k) is `(a)/(2x)+(b)/(2y)=1` or, `(a)/(x)+(b)/(y)=2`