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 November 11th, 2009, 09:34 PM #1 Newbie   Joined: Oct 2009 Posts: 7 Thanks: 0 Please Help me To Findout the Parameter if this circle equation number 1 is given equation equation number 2 is formula please solve this equation answer should be 2(pi)a
 November 11th, 2009, 11:55 PM #2 Newbie   Joined: Nov 2009 Posts: 1 Thanks: 0 Re: Please Help me To Findout the Parameter if this circle The diagram given gives me the impression that the center of the circle is at (0,0). This leads me to assume that equation number 1 is the equation of the circle. Thus the circle has radius a. The parameter/circumfrence of the circle is always 2(pi)r or (pi)d, if the radius is a then the parameter is 2(pi)a as required. Hope it helped
 November 12th, 2009, 12:23 AM #3 Newbie   Joined: Oct 2009 Posts: 7 Thanks: 0 Re: Please Help me To Findout the Parameter if this circle yes u know this thing but my question is how to solve this prob
 November 12th, 2009, 03:40 AM #4 Senior Member   Joined: Mar 2009 Posts: 318 Thanks: 0 You've posted a picture and two equations. What, exactly, is the "question"? Please be complete. Thank you!
 November 12th, 2009, 07:50 AM #5 Newbie   Joined: Oct 2009 Posts: 7 Thanks: 0 Re: Please Help me To Findout the Parameter if this circle i want to findout entire circle length by this forum. by using equation number 2
 November 12th, 2009, 04:34 PM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,943 Thanks: 1134 Math Focus: Elementary mathematics and beyond Re: Please Help me To Findout the Parameter if this circle We'll integrate from  to $a$ (so $y$ is positive) and multiply by 4 to get the circle's circumference. $4\int_0^a\,sqrt{1\,+\,\left(\frac{dy}{dx}\right)^2 }\,dx$ $x^2\,+\,y^2\,=\,a^2,\,y\,=\,\sqrt{a^2\,-\,x^2},\,\frac{dy}{dx}\,=\,-\frac{x}{\sqrt{a^2\,-\,x^2}}$ \begin{align*}4\int_0^a\,sqrt{1\,+\,\left(-\frac{x}{\sqrt{a^2\,-\,x^2}}\right)^2}\,dx\,&=\,4\int_0^a\,sqrt{1\,+\,\ frac{x^2}{a^2\,-\,x^2}}\,dx\\ &=\,4\int_0^a\,\sqrt{\frac{a^2}{a^2\,-\,x^2}}\,dx\\ &=\,4a\int_0^a\,\frac{1}{\sqrt{a^2\,-\,x^2}}\,dx\end{align*} $x\,=\,(a)\sin{\theta},\,dx\,=\,(a)\cos{\theta}\,d\ theta,\,\theta\,=\,\sin^{-1}\left(\frac{x}{a}\right),\,\int\,\frac{1}{\sqrt{ a^2\,-\,x^2}}\,dx\,=\,\int\,\frac{(a)\cos{\theta}}{\sqrt {a^2\,-\,a^2\sin^2{\theta}}}\,d\theta\,=\,\sin^{-1}\left(\frac{x}{a}\right)\,+\,C$ $4a\int_0^a\,\frac{1}{\sqrt{a^2\,-\,x^2}}\,dx\,=\,4a\left(\sin^{-1}(1)\,-\,\sin^{-1}(0)\right)\,=\,2a\pi$ See this: Trigonometric substitution

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