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:
given $\frac{d y}{d x}=\frac{2 y}{x^{2}}$
$\Rightarrow \int \frac{d y}{2 y}=\int \frac{d x}{x^{2}}$
$\Rightarrow \frac{1}{2} \ell \mathrm{ny}=-\frac{1}{\mathrm{x}}+\mathrm{c}$
passes through centre $(1,1)$
$\Rightarrow \mathrm{c}=1$
$\Rightarrow \mathrm{x} \ell \mathrm{ny}=2(\mathrm{x}-1)$