Given that the slope of the tangent to a curve y = y(x) at any point (x,y) is `(2y)/x^(2)` If the surve passes through the ltbegt centre of the circle `x^(2) +y^(2) - 2x -2y =0`, then its equation is (a) `x^(2)log_(e)absy =-2(x-1)` (b) `xlog_(e)absy = x-1` (c) `xlog_(e)absy =2(x-1)` (d) `xlog_(e)absy =-2(x-1)`

Options

A. (a) `x^(2)log_(e)absy =-2(x-1)`
B. (b) `xlog_(e)absy = x-1`
C. (c) `xlog_(e)absy =2(x-1)`
D. (d) `xlog_(e)absy =-2(x-1)`

Answer

(c)

Explanation

Given, `dy/dx= (2y)/x^(2)`
`rArr int dy/y= int (2)/x^(2)dx` [integrating both sides]
`rArr log_(e abs(y))=-2/x+C …(i)`
Since, curve (i) passes through centre (1,1) of the circle
`x^(2)+y^(2)-2x-2y=0`
`therefore log_(e)(1)=-2/1+CrArr C=2`
`therefore` Equation required curve is
`log_(e) abs(y)=2/x+2` [put C=2 in Eq. (i)]
`rArr xlog_(e) abs(y)=2(x-1)`