The area bounded by the circles \(\mathrm{x}^{2}+\mathrm{y}^{2}=1, \mathrm{x}^{2}+\mathrm{y}^{2}=4\) and the pair of lines \(\sqrt{3}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=4 \mathrm{xy}\) is equal to \(\ldots \ldots \ldots\) (a) \((\pi / 4)\) (b) \((\pi / 2)\) (c) \((5 / 2)\) (d) 3

Given two circles with equations: 1. \(x^2 + y^2 = 1\) (Circle 1) 2. \(x^2 + y^2 = 4\) (Circle 2) Circle 1 is centered at the origin (0,0) with radius 1, and Circle 2 is also centered at the origin with radius 2.

Given the pair of lines equation: \(\sqrt{3}(x^2 + y^2) = 4xy\) Rewrite the equation in terms of y to get: \(y = \frac{\sqrt{3}(x^2 + y^2)}{4x}\) \(\Rightarrow y^2 = \frac{3(x^2 + y^2)}{16x^2}\) \(\Rightarrow y^2 = \frac{3x^2 + 3y^2}{16x^2}\) \(\Rightarrow 16x^2y^2 = 3x^2(13y^2 - 3x^2)\) Since \(x^2 \neq 0\), we can divide both sides by \(x^2\), which simplifies the equation to: \(16y^2 = 3(13y^2 - 3x^2)\) Now, we will find the points of intersection between the circle and the lines.

To find the points of intersection between the circles and lines, we will substitute one equation into the other. From the equation of Circle 1, we have: \(y = \sqrt{1 - x^2}\) Plug this into the line equation to get: \(16(1 - x^2) = 3(13 - 3x^2)\) Simplifying, we get: \(16 = 39x^2\) \(\Rightarrow x^2 = \frac{16}{39}\) Now, let's find the corresponding y-coordinates by plugging back into the Circle 1 equation: \(y^2 = 1 - x^2 = 1 - \frac{16}{39} = \frac{23}{39}\) \(\Rightarrow y = \pm \sqrt{\frac{23}{39}}\) Hence, the points of intersection are \(\left(\pm \sqrt{\frac{16}{39}}, \pm \sqrt{\frac{23}{39}}\right)\).

To find the area enclosed by the circles and the lines, we will integrate the difference between their radial functions over the angles for each region. For Circle 1, the angle swept out by the lines is given by: \(\alpha = 2 \arctan \frac{\sqrt{23}}{4}\), then the area enclosed by Circle 1 and lines is given by: \(A_1 = \int_{-\alpha}^{\alpha} \frac{1}{2} r_1^2 d\theta = \int_{-\alpha}^{\alpha} \frac{1}{2} d\theta\) For Circle 2, the angle swept out by the lines is given by: \(\beta = 2 \arctan \frac{\sqrt{23}}{16}\), then the area enclosed by Circle 2 and lines is given by: \(A_2 = \int_{-\alpha}^{\alpha} \frac{1}{2} r_2^2 d\theta = \int_{-\alpha}^{\alpha} 2 d\theta\) Thus, the total area enclosed (\(A_{total}\)) is given by: \(A_{total} = A_2 - A_1\)

Evaluating the integrals, we get: \(A_{total} = \left(2 \times 2\arctan\frac{\sqrt{23}}{16} - \frac{1}{2} \times 2\arctan\frac{\sqrt{23}}{4}\right) = \frac{3}{2} \times 2\arctan\frac{\sqrt{23}}{4}\) Since the options are given in terms of \(\pi\), we can find a relationship between the angle \(\alpha\) and \(\pi\). We know that \(\tan\alpha = \frac{\sqrt{23}}{4}\), so \(\sin\alpha = \frac{\sqrt{23}}{\sqrt{401}}\) and \(\cos\alpha = \frac{4}{\sqrt{401}}\). So, the half-angle formula for \(\tan\) gives us: \(\tan\frac{\alpha}{2} = \frac{\sin\alpha}{1 + \cos\alpha}\) Plugging in the values, we get: \(\tan\frac{\alpha}{2} = \frac{\sqrt{3}}{3}\) Now we can rewrite our integral in terms of \(\pi\), and evaluate the total area: \(A_{total} = \frac{3}{2}(2\arctan\frac{\sqrt{23}}{4}) = \frac{3}{2}(2\frac{\pi}{6})\) \(A_{total} = \frac{3}{2}(\frac{\pi}{3}) = \frac{1}{2}\pi\) So our answer is (b) \(\frac{1}{2}\pi\).