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…

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
  1. $\quad x \log |y|=x-1$
  2. $x \log |y|=-2(x-1)$
  3. $\quad x \log |y|=2(x-1)$
  4. $x^2 \log |y|=-2(x-1)$

Solution

$\begin{aligned} & \text { Equation of the given circle is } \\ & x^2+y^2-2 x-2 y=0 \\ & \therefore \quad x^2-2 x+1+y^2-2 y+1=2 \\ & \therefore \quad(x-1)^2+(y-1)^2=2 \\ & \therefore \quad \text { Centre of the circle is }(1,1) \end{aligned}$
Now, slope of the given tangent is $\frac{2 y}{x^2}$ i.e., $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2 y}{x^2}$. $\therefore \quad$ Integrating on both sides, we get $\int \frac{1}{y} \mathrm{~d} y=2 \int x^{-2} \mathrm{~d} x$ $\therefore \quad \log |y|=\frac{-2}{x}+c$...(i) At $(1,1)$, we get $\log 1=-2+c$ $\therefore \quad \mathrm{c}=2$ $\therefore \quad$ Required equation is $\begin{aligned} & \log |y|=\frac{-2}{x}+2 \\ & \text { i.e., } x \log |y|=2(x-1)\end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 1)

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