Question

Find the area of the region enclosed by the circles (x-1)^2 +y^2 = 1 and x^2 +y^2 = 1​

Answer 1

Answer:

Area of enclosed region = $\frac{2\pi}{3}-\frac{\sqrt{3}}{2}$ sq.units.

Step-by-step explanation:

Given :

To find the area of the region enclosed by the circles

$(x-1)^2 +y^2 = 1$ &

$x^2 +y^2 = 1$.

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Solution :

let the equations,

$(x-1)^2 +y^2 = 1$ ...(i)

$x^2 +y^2 = 1$ ...(ii)

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We need to know the point of intersection , to find the region enclosed,.

i.e., both equations should posses two equal values of x & y .

Which can be found by equating them,.

$(x-1)^2 +y^2 = 1$

$x^2 +y^2 = 1$

⇒ $(x-1)^2 +y^2 = 1 = x^2 +y^2$

⇒ $(x-1)^2 +y^2 = x^2 + y^2$

⇒ $(x-1)^2 = x^2$

⇒ $x^2 - 2x + 1 = x^2$

⇒ $-2x + 1 = 0$

⇒ $x = \frac{1}{2}$

by substituting value of x in (i),

We get,

$(x-1)^2 +y^2 = 1$

$(\frac{1}{2}-1)^2 +y^2 = 1$

$(\frac{-1}{2})^2 +y^2 = 1$

$\frac{1}{4}+y^2 = 1$

$y^2 = 1-\frac{1}{4}=\frac{3}{4}$

$y = \frac{\sqrt{3}}{2}$ (or) $y = -\frac{\sqrt{3}}{2}$

Hence, the points are,

$(\frac{1}{2},\frac{\sqrt{3}}{2})$ & $(\frac{1}{2},-\frac{\sqrt{3}}{2})$.

so, the area must lie in circular phase between these points,.

refer to the graph attached,.

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The curve from x = 0 to x = $\frac{1}{2}$,.

must be from the equation,

$(x-1)^2 +y^2 = 1$

⇒ $y^2 = 1-(x-1)^2$

⇒ $y = \sqrt{1-(x-1)^2}$

So, the curve, $y = \sqrt{1-(x-1)^2}$

moves from x=0 to x= $\frac{1}{2}$.

The curve from x = $\frac{1}{2}$ to x = 1,.

must be from the equation,

$x^2 +y^2 = 1$

⇒ $y^2 = 1-x^2$

⇒ $y = \sqrt{1-x^2}$

So, the curve, $y = \sqrt{1-x^2}$

moves from x = $\frac{1}{2}$ to x = 1.

Hence,.

The area is symmetrical about x-axis (refer to the graph),

$A = 2 \times [(\int\limits^\frac{1}{2}_0 {\sqrt{1-(x-1)^2}} \, dx ) + (\int\limits^1_\frac{1}{2} {\sqrt{1-x^2}} \, dx )$

⇒ $A = 2 \times [\frac{1}{2}. (x-1).\sqrt{1-(x-1)^2}+sin^{-1}(x-1)]^{\frac{1}{2} }_0 + [\frac{1}{2}.x.\sqrt{1-x^2} + sin^{-1}(x)]^1_{\frac{1}{2}}$

⇒ $A = 2 \times [\frac{1}{2}. ((\frac{1}{2}-0)-1).\sqrt{1-((\frac{1}{2}-0)-1)^2}+sin^{-1}(\frac{1}{2}-1)-sin^{-1}(0-1)]+ [\frac{1}{2}.(1-\frac{1}{2}).\sqrt{1-(1-\frac{1}{2})^2} + sin^{-1}(1) - sin^{-1}(\frac{1}{2})]$

⇒ $-\frac{\sqrt{3}}{4} - \frac{\pi}{6} +\frac{\pi}{2}+\frac{\pi}{2}-\frac{\sqrt{3}}{4}-\frac{\pi}{6} = \frac{2\pi}{3}-\frac{\sqrt{3}}{2}$

∴ Area of enclosed region = $\frac{2\pi}{3}-\frac{\sqrt{3}}{2}$ sq.units.