Question

The area of the region lying inside $x^2+(y-1{)}^2=1$ and out side $c^2{x}^2+y^2=c^2,$ where $c=(\sqrt{2}-1)$

• A. $(4-\sqrt{2})\frac{\pi }{4}+\frac{1}{\sqrt{2}}$
• B. $(4+\sqrt{2})\frac{\pi }{4}-\frac{1}{\sqrt{2}}$
• C. $(4+\sqrt{2})\frac{\pi }{4}+\frac{i}{\sqrt{2}}$
• D. None of these

Answer 1

Answer: (A)

$(4-\sqrt{2})\frac{\pi }{4}+\frac{1}{\sqrt{2}}$

$x^2+{(y-1)}^2=1........\left(i\right)$
$c^2{x}^2+y^2=c^2...........\left(ii\right)$
$wherec=(\sqrt{2}-1)$
$solving\left(i\right)and\left(ii\right)$
${(y-1)}^2=\frac{y^2}{c^2}$
$\Rightarrow (\frac{y}{c}+y-1)(\frac{y}{c}-y+1)=0$
$\Rightarrow y=\frac{c}{1+c},\frac{c}{c-1}$
$Now$
$x^2=1-\frac{y^2}{c^2}$
$\Rightarrow x^2=1-\frac{1}{{(c+1)}^2}$
$=1-\frac{1}{{\left(\sqrt{2}\right)}^2}$
$=\frac{1}{2}$
$Hencex=\pm \frac{1}{\sqrt{2}}$
$AreaofOABCO$
$2[\underset{0}{\overset{1/\sqrt{2}}{\int }}c\sqrt{1-x^2}dx-\underset{0}{\overset{1/\sqrt{2}}{\int }}(1-\sqrt{1-x^2})dx]$
$=2[\underset{0}{\overset{1/\sqrt{2}}{\int }}(c+1)\sqrt{1-x^2}dx-\underset{0}{\overset{1/\sqrt{2}}{\int }}\left(1\right)dx]$
$=2[\sqrt{2}\underset{0}{\overset{1/\sqrt{2}}{\int }}\sqrt{1-x^2}dx-\frac{1}{\sqrt{2}}]$
$=2[\frac{\sqrt{2}}{2}{(x\sqrt{1-x^2}+{\sin }^{-1}x)}_0^{1/\sqrt{2}}-\frac{1}{\sqrt{2}}]$
$=2(\frac{\sqrt{2}}{2}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}+\frac{\sqrt{2}}{2}\frac{\pi }{4}\frac{1}{\sqrt{2}})$
$=\frac{\sqrt{2}\pi }{4}-\frac{\sqrt{2}}{2}$
$Hencetherequiredarea$
$\pi -\frac{\sqrt{2}\pi }{4}-\frac{\sqrt{2}}{2}$
$=(4-\sqrt{2})\frac{\pi }{4}+\frac{1}{\sqrt{2}}$