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 August 16th, 2013, 04:32 PM #1 Member   Joined: Aug 2013 Posts: 59 Thanks: 0 Circle If there's a circle x² + y² = r² (r > 0) and the line tangent to such circle has the equation y = a, determine the two real number answers for a.
 August 16th, 2013, 05:28 PM #2 Member   Joined: Aug 2013 Posts: 40 Thanks: 3 Re: Circle since y=a is horizontal, the coordinates of contact will be (0, r) and /(0, -r). Thus a=r and a=-r
 August 16th, 2013, 07:52 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 Eliminating y from the given equations, one gets x² + a² - r² = 0. This equation must have "equal roots" (zero discriminant), so a = ±r.
 August 16th, 2013, 09:49 PM #4 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234 Re: Circle You can also let x = 0 from which you get y = ±r = a immediately.
 August 16th, 2013, 11:16 PM #5 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 Without explaining why you take x = 0?
 August 16th, 2013, 11:34 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Re: Circle Wonder why Slipper posted that problem...quite obvious that a circle radius r centered at origin has 2 tangent points (to horizontal lines) at (0,r) and (0,-r).
August 17th, 2013, 01:07 AM   #7
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Joined: Jul 2011
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Re:

Quote:
 Originally Posted by skipjack Without explaining why you take x = 0?
I had a hunch you would say that. I guess we can plug in the constant 'a' , take the derivative , set it to zero to confirm the horizontal tangents indeed occur when x = 0 , then go back and let x = 0 to get y = ±r = a.

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