Question

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 :

• A. \(\frac{\pi}{4} + 1\)
• B. \(\pi + 1\)
• C. \(\pi - 1\)
• D. \(\frac{\pi}{2} - 1\)

Hint:

Both curves turn out to be unit circles; the required area is their common (lens) region.

Answer 1

The Correct Option is D

\(C_1\): put \(v = y^2\): \(x\frac{dv}{dx} = v - x^2\Rightarrow \frac{d}{dx}\left(\frac{v}{x}\right) = \frac{x v' - v}{x^2} = -1\), so \(\frac{y^2}{x} = -x + c\), i.e. \(x^2 + y^2 = cx\). Through \((1,1)\): \(c = 2\), so \(C_1: x^2 + y^2 = 2x\), a circle with centre \((1,0)\) and radius 1.
\(C_2\): \(\frac{dx}{dy} = \frac{x^2 - y^2}{2xy}\). With \(u = x^2\): \(y\frac{du}{dy} = u - y^2\Rightarrow \frac{d}{dy}\left(\frac{u}{y}\right) = -1\), so \(x^2 + y^2 = c'y\). Through \((1,1)\): \(c' = 2\), so \(C_2: x^2 + y^2 = 2y\), a circle with centre \((0,1)\) and radius 1.
The circles intersect at \((0,0)\) and \((1,1)\); the common chord has length \(\sqrt{2}\) and subtends \(90^\circ\) at each centre.
Area of lens \(= 2\left(\text{sector} - \text{triangle}\right) = 2\left(\frac{\pi}{4} - \frac{1}{2}\right) = \frac{\pi}{2} - 1\).
Answer: (4) \(\frac{\pi}{2} - 1\)