February 9th, 2015, 01:32 PM  #1 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty  Cartesian Circle
Find the equation of the circle. I'm given three sets of coordinates through which the circle passes through: 1. $\displaystyle (3, 0)$ 2. $\displaystyle (0, 9)$ 3. $\displaystyle (0, 1)$ $\displaystyle r^2 = (x  K)^2 + (y  H)^2$ (3, 0): $\displaystyle r^2 = (3  K)^2 + (0  H)^2$ > $\displaystyle r^2 = (3  K)^2 + H^2$ [1] (0, 9): $\displaystyle r^2 = (0  K)^2 + (9  H)^2$ > $\displaystyle r^2 = K^2 + (9  H)^2$ [2] (0, 1): $\displaystyle r^2 = (0  K)^2 + (1  H)^2$ > $\displaystyle r^2 = K^2 + (1  H)^2$ [3] I substitute [2] into [3] $\displaystyle K^2 + (1  H)^2 = K^2 + (9  H)^2$ $\displaystyle (1  H)^2  (9  H)^2 = K^2  K^2$ $\displaystyle (1  H)^2  (9  H)^2 = 0$ $\displaystyle H^2  2H + 1  H^2  18H + 81 = 0$ $\displaystyle 2H  18H + 1 + 81 = 0$ $\displaystyle 20H + 82 = 0$ $\displaystyle H = 82/20$ $\displaystyle H = 4.1$ substituting [1] into [2] i get $\displaystyle K  3H + 12 = 0$ H = 4.1 so K = 0.3 I GOT THE ABOVE QUESTION WRONG. COULD SOMEONE SHOW ME WHERE I HAVE GONE WRONG PLEASE. THANK YOU. 
February 9th, 2015, 02:34 PM  #2 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244  
February 9th, 2015, 04:44 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, hyperbola! You made some algebraic errors. Quote:
[1] [2] [3] Equate [2] and [3] Substitute into [1]: Substitute into [2]: Equate: Substitute into [1]: Equation:  
February 9th, 2015, 08:34 PM  #4 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty 
Thanks. Went back and checked and yes a simple algebraic mistake I'll have to work on not rushing through question(s) 

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cartesian, circle 
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