BITSAT 2025 Shift 2 Paper

Given circles are $x^{2}+y^{2}=1$and $(x-1)^{2}+y^{2}=1$Centre of (i) is $O(0,0)$ and radius $=1$Both these circle are symmetrical about x -axis solving (i) and (ii), we get, $-2x+1=0 \Rightarrow x=\;\frac{1}{2}$then $y^{2}=1- \left(\;\frac{1}{2}\right)^{2}=\;\frac{3}{4} \Rightarrow y=\;\frac{\sqrt{3}}{2}$∴ The points of intersection are$P\left(\;\frac{1}{2},\;\frac{\sqrt{3}}{2}\right)$ and $Q\left(\;\frac{1}{2},-\;\frac{\sqrt{3}}{2}\right)$It is clear from the figure that the shaded portion in region whose area is required.∴ Required area $=$ area OQAPO$=2 \times$ area of the region OLAP$=2 \times ($ area of the region OLPO + area of LAPL $)$$=2 \left[ \int\limits_{0}^{\frac{1}{2}} \sqrt{1-(x-1)^{2}} \, dx + \int\limits_{\frac{1}{2}}^{1} \sqrt{1-x^{2}} \, dx \right]$$=2 \left[ \;\frac{(x-1)\sqrt{1-(x-1)^{2}}}{2} + \;\frac{1}{2} \sin^{-1}(x-1) \right]_{0}^{\frac{1}{2}}$$\; \;\; +2 \left[ \;\frac{x\sqrt{1-x^{2}}}{2} + \;\frac{1}{2} \sin^{-1} x \right]_{\frac{1}{2}}^{1}$$\; = -\;\frac{1}{2} \cdot \;\frac{\sqrt{3}}{2} + \sin^{-1} \left( \;\frac{-1}{2} \right) - \sin^{-1}(-1) + 0$$\; \;\; + \sin^{-1}(1) - \left( \;\frac{1}{2} \cdot \;\frac{\sqrt{3}}{2} + \sin^{-1} \left( \;\frac{1}{2} \right) \right)$$\; = \left( \;\frac{2\pi}{3} - \;\frac{\sqrt{3}}{2} \right) \;\text{ sq. units. }\;$