Given that the slope of the tangent to a curve y=y(x) at any point (x,y) is 2yx2. If the curve passes through the centre of the circle x2+y2−2x−2y=0, then its equation is :
A
xloge|y|=2(x−1)
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B
xloge|y|=(x−1)
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C
xloge|y|=−2(x−1)
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D
x2loge|y|=−2(x−1)
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Solution
The correct option is Axloge|y|=2(x−1) dydx=2yx2 ⇒dyy=2dxx2 Integrating both sides ∫dyy=∫2dxx2 ⇒loge|y|=−2x+c
The curve passes through the centre of the circle x2+y2−2x−2y=0, i.e., (1,1) ⇒0=−2+c ⇒c=2 ∴ Equation of the curve is loge|y|=−2x+2 ⇒xloge|y|=2(x−1)