A circle passes through a fixed point A and cuts two perpendicular straight lines through A in B and C. If the straight line BC passes through a fixed-point `(x_(1), y_(1))`, the locus of the centre of the circle, is

Options

A. `(x_(1))/(x)+(y_(1))/(y)=1`
B. `x_(1)y=x_(1)y_(1)`
C. `xy_(1)+yx_(1)=2`
D. `(x_(1))/(x)+(y_(1))/(y)=2`

Answer

D

Explanation

Let A be the origin, and let AB and AC be x and y axes respectively, Let the coordinates of B and C be (a, 0) and (0, b) respectively. Then, equation of BC is
`(x)/(a)+(y)/(b)=1 " " ...(i)`
Let (h, k) be the coordinates of the centre of the circle of the circle. Clearly, BC is a diameter of the circle.

`h=(a)/(2)` and `k=(b)/(2)rArr a=2h` and `b=2k`.
Since (i) passes through `(x_(1), y_(1))`
`:. (x_(1))/(a) + (y_(1))/(b)=1 rArr (x_(1))/(2h)+(y_(1))/(2k)=1 " " [:' a=2h, b=2k]`
Hence, the locus of (h, k) is `(x_(1))/(2x)+(y_(1))/(2y)=1` or, `(x_(1))/(x)+(y_(1))/(y)=2`