Given that the slope of the tangent to a curve \(y=y(x)\) at any point \((x, y)\) is \(\frac{2 y}{x^2}\). If the curve passes through the centre of the circle \(x^2+y^2-2 x-2 y=0\), then its equation is

\(\begin{aligned}
& \text { Equation of the given circle is } \\
& x^2+y^2-2 x-2 y=0 \\
& \therefore \quad x^2-2 x+1+y^2-2 y+1=2 \\
& \therefore \quad(x-1)^2+(y-1)^2=2 \\
& \therefore \quad \text { Centre of the circle is }(1,1)
\end{aligned}\)
Now, slope of the given tangent is \(\frac{2 y}{x^2}\)
i.e., \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2 y}{x^2}\).
\(\therefore \quad\) Integrating on both sides, we get
\(\int \frac{1}{y} \mathrm{~d} y=2 \int x^{-2} \mathrm{~d} x\)
\(\therefore \quad \log |y|=\frac{-2}{x}+c\)...(i)
At \((1,1)\), we get
\(\log 1=-2+c\)
\(\therefore \quad \mathrm{c}=2\)
\(\therefore \quad\) Required equation is
\(\begin{aligned} & \log |y|=\frac{-2}{x}+2 \\ & \text { i.e., } x \log |y|=2(x-1)\end{aligned}\)