The area (in square units) of the region encl | vedclass

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(D) The intersection points of the two circles $x^2+y^2=1$ $(i)$ and $(x-1)^2+y^2=1$ (ii) are found by substituting $y^2=1-x^2$ from $(i)$ into (ii):$

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$\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}$

Vedclass · Answer

(D) The intersection points of the two circles $x^2+y^2=1$ $(i)$ and $(x-1)^2+y^2=1$ (ii) are found by substituting $y^2=1-x^2$ from $(i)$ into (ii):$(x-1)^2+(1-x^2)=1$$x^2-2x+1+1-x^2=1$$-2x+2=1 \Rightarrow x=\frac{1}{2}$Substituting $x=\frac{1}{2}$ into $(i)$, we get $y^2=1-\frac{1}{4}=\frac{3}{4} \Rightarrow y=\pm\frac{\sqrt{3}}{2}$.The intersection points are $A\left(\frac{1}{2}, \frac{\sqrt{3}}{2}ight)$ and $C\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}ight)$.The area of the enclosed region is symmetric about the $x$-axis, so the total area is $2 	imes$ (Area of the region in the first quadrant bounded by $x=0, x=\frac{1}{2}, x=1$ and the arcs of the circles).Area $= 2 \left[ \int_0^{1/2} \sqrt{1-(x-1)^2} dx + \int_{1/2}^1 \sqrt{1-x^2} dx ight]$Using the formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$:Area $= 2 \left[ \left( \frac{x-1}{2}\sqrt{1-(x-1)^2} + \frac{1}{2}\sin^{-1}(x-1) ight)_0^{1/2} + \left( \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1}(x) ight)_{1/2}^1 ight]$$= 2 \left[ \left( \frac{-1}{4}\frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}(-\frac{1}{2}) - (0 + \frac{1}{2}\sin^{-1}(-1)) ight) + \left( (0 + \frac{1}{2}\sin^{-1}(1)) - (\frac{1}{4}\frac{\sqrt{3}}{2} + \frac{1}{2}\sin^{-1}(\frac{1}{2})) ight) ight]$$= 2 \left[ -\frac{\sqrt{3}}{8} - \frac{\pi}{12} + \frac{\pi}{4} + \frac{\pi}{4} - \frac{\sqrt{3}}{8} - \frac{\pi}{12} ight]$$= 2 \left[ \frac{2\pi}{3} - \frac{\sqrt{3}}{4} ight] = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$.

The area (in square units) of the region enclosed by the two circles $x^2+y^2=1$ and $(x-1)^2+y^2=1$ is

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