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(D) The slope of the tangent is given by $\frac{dy}{dx} = \frac{2y}{x^2}$.Separating the variables,we get $\frac{dy}{y} = \frac{2}{x^2} dx$.Integratin

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

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(D) The slope of the tangent is given by $\frac{dy}{dx} = \frac{2y}{x^2}$.Separating the variables,we get $\frac{dy}{y} = \frac{2}{x^2} dx$.Integrating both sides,$\int \frac{dy}{y} = \int 2x^{-2} dx$,which gives $\ln |y| = -\frac{2}{x} + C$.The circle equation is $x^2 + y^2 - 2x - 2y = 0$,which can be written as $(x-1)^2 + (y-1)^2 = 2$. The centre is $(1, 1)$.Since the curve passes through $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the equation: $\ln |1| = -\frac{2}{1} + C$,so $0 = -2 + C$,which implies $C = 2$.Thus,the equation is $\ln |y| = -\frac{2}{x} + 2$.Multiplying by $x$,we get $x \ln |y| = -2 + 2x = 2(x - 1)$.

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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