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Given that the slope of the tangent to a curve \( y=y(x) \) at
P any point \( (x, y) \) is \( \frac{2 y}{x^{2}} \). If the curve passes through the

W) centre of the circle \( x^{2}+y^{2}-2 x-2 y=0 \), then its equation is:
[JEE Main - 2019(April)]
(a) \( x \log _{\mathrm{e}}|y|=2(x-1) \)
(b) \( x \log _{e}|y|=2(x-1) \)
(c) \( x^{2} \log _{e}|y|=-2(x-1) \)
(d) \( x \log _{e}|y|=-2(x-1) \)
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