Question:
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 :
Correct Option: 1,
Solution:
Given $\frac{d y}{d x}=\frac{2 y}{x^{2}}$
Integrating both sides, $\int \frac{d y}{y}=2 \int \frac{d x}{x^{2}}$
$\Rightarrow \ln |y|=-\frac{2}{x}+C$ .........(1)
Equation (i) passes through the centre of the circle
$x^{2}+y^{2}-2 x-2 y=0$, i.e., $(1,1)$
$\therefore C=2$
Now, $\ln |y|=-\frac{2}{x}+2$
$x \ln |y|=-2(1-x) \Rightarrow x \ln |y|=2(x-1)$
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