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 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 co-ordinates 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
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Quote:
 Originally Posted by hyperbola Find the equation of the circle. $\displaystyle (1 - H)^2 - (9 - H)^2 = 0$ $\displaystyle H^2 - 2H + 1 - H^2 - 18H + 81 = 0$
You've expanded this slightly wrong.$$(1 - H)^2 - (9 - H)^2 = H^2 - 2H + 1 - (H^2 - 18H + 81) = H^2 - 2H + 1 - H^2 + 18H - 81$$

February 9th, 2015, 04:44 PM   #3
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Hello, hyperbola!

Quote:
 Find the equation of the circle. I'm given three sets of co-ordinates through which the circle passes through: $\;\;\;(3,\,0),\;\;(0,\,9),\;\;(0,\,1)$
$r^2 \:=\: (x\,- \,k)^2 + (y\,-\,h)^2$

$(3,\,0): \;\;r^2 \:=\: (3\,-\,k)^2 + (0\,-\,h)^2 \;\;\;\Rightarrow\;\;\;r^2 \:=\: (3\,-\,k)^2\,+\,h^2\;\;\;$ [1]

$(0,\,9):\;\; r^2 \:=\: (0\,-\,k)^2\,+\,(9\,-\,h)^2\;\;\;\Rightarrow\;\;\; r^2 \:=\: k^2\,+\,(9\,-\,h)^2\;\;$ [2]

$(0,\,1):\;\;r^2 \:=\: (0\,-\,k)^2\,+\,(1\,-\,h)^2\;\;\;\Rightarrow \;\;\; r^2 \:=\: k^2\,+\,(1\,-\,h)^2\;\;$[3]

Equate [2] and [3]

$\;\;\;k^2\,+\,(1\,-\,h)^2 \:=\: k^2\,+\,(9\,-\,h)^2$

$\;\;\;(1\,-\,h)^2\,-\,(9\,-\,h)^2 \:=\:0$

$\;\;\;(1\,-\,2h\,+\,h^2)\,-\,(81\,-\,18h\,+\,h^2) \:=\:0$

$\;\;\;1\,-\,2h\,+\,h^2\,-\,81\,+\,18h\,-\,h^2 \:=\:0$

$\;\;\;16h\,-\,80 \:=\:0$

$\;\;\;\boxed{h \,=\,5}$

Substitute into [1]: $\: r^2 \:=\: (3\,-\,k)^2 + 5^2$
Substitute into [2]: $\:r^2 \:=\: k^2\,+\,4^2$

Equate: $\: (3\,-\,k)^2\,+\,25 \:=\:k^2\,+\,16$

$\;\;\;\;\;9\,-\,6k\,+\,k^2\,+\,25 \:=\: k^2\,+\,16$

$\;\;\;\;\;-6k \:=\:-18 \;\;\;\Rightarrow\;\;\;\boxed{k \,=\,3}$

Substitute into [1]: $\;r^2 \:= \: (3\,-\,3)^2\,+\,5^2 \;\;\;\Rightarrow\;\;\;\boxed{r^2 \,=\,25}$

Equation: $\:(x\,-\,3)^2\,+\,(y\,-\,5)^2 \:=\:25$

 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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