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 October 5th, 2013, 11:03 PM #1 Member   Joined: Aug 2013 Posts: 69 Thanks: 0 translating a circle Hi, I want to translate y = sqrt.(5x-x^2) down 2 and to the left 7. Putting the above equation into standard form, I get y^2 +(x-5/2)^2 = 25/4 (y+2)^2 +(x+9/2)^2= 25/4 How do I get this into y=x form? Brain is dead :+}
October 6th, 2013, 04:18 AM   #2
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Re: translating a circle

Quote:
 Originally Posted by yogazen2013 I want to translate y = sqrt.(5x-x^2) down 2 and to the left 7.
$y\,=\,\sqrt{5(x\,+\,7)\,-\,(x\,+\,7)^2}\,-\,2$

 October 6th, 2013, 10:54 AM #3 Member   Joined: Aug 2013 Posts: 69 Thanks: 0 Re: translating a circle Can you show me the steps? And I this circle translated 7 from the original point, so iy is @ -4.5=x
 October 6th, 2013, 02:36 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 Translating y = f(x) down by 2 and to the left by 7 gives y = f(x + 7) - 2.
October 7th, 2013, 06:53 AM   #5
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Re: translating a circle

Quote:
 Originally Posted by yogazen2013 Can you show me the steps? And I this circle translated 7 from the original point, so iy is @ -4.5=x
Solve the equation $(y+2)^2 +(x+9/2)^2= 25/4$ for y.

Subtract $(x+ 9/2)^2$ from both sides:
$(y+ 2)^2= 25/4- (x+ 9/2)^2$

Take the square root of both sides:
$y+ 2= \pm\sqrt{25/4- (x+ 9/2)^2}$

Subtract 2 from both sides:
$y= -2\pm\sqrt{25/4- (x+ 9/2)^2}$

You can, if you want, multiply out that $(x+ 9/2)^2= x^2+ 9x+ 81/4$ to get $25/4- (x+9/2)^2= -14- 9x- x^2$ inside the square root.

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