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Given that the slope of the tangent to a curve y = y(x) atany 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 itsequation 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)

Answer»

(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)`

SOLUTION :Given,`dy/DX= (2y)/x^(2)`
`rArrint dy/y= int (2)/x^(2)dx`[integrating both sides]
`rArrlog_(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`
`thereforelog_(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)]
`rArrxlog_(e) abs(y)=2(x-1)`


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