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(C) The equation of the given circle is $x^2+y^2-2x-2y=0$. Completing the square,we get $(x-1)^2+(y-1)^2=2$. Thus,the centre of the circle is $(1, 1)$

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$x \log |y|=2(x-1)$

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(C) The equation of the given circle is $x^2+y^2-2x-2y=0$. Completing the square,we get $(x-1)^2+(y-1)^2=2$. Thus,the centre of the circle is $(1, 1)$. The slope of the tangent to the curve is given by $\frac{dy}{dx} = \frac{2y}{x^2}$. Separating the variables,we have $\int \frac{1}{y} dy = \int \frac{2}{x^2} dx$. Integrating both sides,we get $\log |y| = -\frac{2}{x} + c$. Since the curve passes through $(1, 1)$,we substitute $x=1$ and $y=1$: $\log |1| = -\frac{2}{1} + c \implies 0 = -2 + c \implies c = 2$. Substituting $c=2$ into the general solution,we get $\log |y| = -\frac{2}{x} + 2$. Multiplying by $x$,we get $x \log |y| = -2 + 2x = 2(x-1)$.

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

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