Given that the slope of the tangent to a curve
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

1. $x \log _{e}|y|=2(x-1)$

2. $x \log _{e}|y|=x-1$

3. $x^{2} \log _{e}|y|=-2(x-1)$

4. $x \log _{e}|y|=-2(x-1)$

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 $(1,1)$

$\Rightarrow \mathrm{c}=1$

$\Rightarrow \mathrm{x} \ell \mathrm{ny}=2(\mathrm{x}-1)$