JEE Main 2021 (Online) 16th March Evening ShiftQuestion 1
Let $C_1$ be the curve obtained by the solution of differential equation $2xy\frac{dy}{dx}=y^{2}-x^{2}$, $x>0$. Let the curve $C_2$ be the solution of $\frac{2xy}{x^{2}-y^{2}}=\frac{dy}{dx}$. If both the curves pass through $(1,1)$, then the area enclosed by the curves $C_1$ and $C_2$ is equal to :
Differential Equation
Solution
Concept. Both equations are homogeneous; put $y=vx$. They give two unit circles, and the area is their common lens. Formulas used. (1) $y=vx\Rightarrow y'=v+xv'$(2) Segment of a circle: $\frac{r^2}{2}(\theta-\sin\theta)$ for central angle $\theta$Step 1 - $C_1$. By (1), $2v(v+xv')=v^2-1$, so $\dfrac{2v\,dv}{1+v^2}=-\dfrac{dx}{x}$ and $x(1+v^2)=c$, i.e. $x^2+y^2=cx$. Through $(1,1)$: $x^2+y^2=2x$ (centre $(1,0)$, radius $1$). Step 2 - $C_2$. By (1), $xv'=\dfrac{v(1+v^2)}{1-v^2}$, and $\dfrac{1-v^2}{v(1+v^2)}=\dfrac1v-\dfrac{2v}{1+v^2}$. So $\dfrac{v}{1+v^2}=cx$, i.e. $x^2+y^2=\frac{y}{c}$. Through $(1,1)$: $x^2+y^2=2y$ (centre $(0,1)$, radius $1$). Step 3 - Lens. The circles meet at $(0,0)$ and $(1,1)$. The chord has length $\sqrt2$, so in each circle it subtends a right angle. By (2), each segment is $\frac12\left(\frac\pi2-1\right)$: $$\text{area}=2\cdot\frac12\left(\frac\pi2-1\right)=\frac\pi2-1.$$ Answer: (C)