My Math Forum Find equation of this circle

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 March 7th, 2010, 10:26 AM #1 Newbie   Joined: Mar 2010 Posts: 2 Thanks: 0 Find equation of this circle Find the equation of the circle centered at (3,0) that intersects the curve y=x^2 at exactly one point.
 March 8th, 2010, 08:56 AM #2 Senior Member   Joined: Apr 2008 Posts: 435 Thanks: 0 Re: Find equation of this circle $(x-3)^2 + y^2 = r^2 \\ x^2 - 6x + y^2 = r^2 \\ 2x - 6 + 2yy' = 0\\ y' = (6-2x)/(2y)\\ But \ also \ y' = 2x \\ So 2x = (6-2x)/(2x^2)\\ And \ then \ it's\ only \ one \ step \ away \ from \ the \ final \ answer\\ \ \ Which \ I'll \ let \ you \ get.$

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