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Question:
Given that the slope of the tangent to a curve $\mathrm{y}=\mathrm{y}(\mathrm{x})$ at any point $(\mathrm{x}, \mathrm{y})$ is $\frac{2 \mathrm{y}}{\mathrm{x}^{2}}$. 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:
$\operatorname{given} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{y}}{\mathrm{x}^{2}}$
$\Rightarrow \int \frac{\mathrm{dy}}{2 \mathrm{y}}=\int \frac{\mathrm{dx}}{\mathrm{x}^{2}}$
$\Rightarrow \frac{1}{2} \ell \mathrm{ny}=-\frac{1}{\mathrm{x}}+\mathrm{c}$
passes through centre $(1,1)$
$\Rightarrow \mathrm{c}=1$
$\Rightarrow \mathrm{x} \ell \mathrm{ny}=2(\mathrm{x}-1)$
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