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February 9th, 2015, 01:32 PM   #1
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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.
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February 9th, 2015, 02:34 PM   #2
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Quote:
Originally Posted by hyperbola View Post
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.$$
(1 - H)^2 - (9 - H)^2 = H^2 - 2H + 1 - (H^2 - 18H + 81) = H^2 - 2H + 1 - H^2 + 18H - 81$$
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February 9th, 2015, 04:44 PM   #3
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Hello, hyperbola!

You made some algebraic errors.


Quote:
Find the equation of the circle.

I'm given three sets of co-ordinates through which the circle passes through:



[1]

[2]

[3]


Equate [2] and [3]














Substitute into [1]:
Substitute into [2]:

Equate:





Substitute into [1]:


Equation:

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February 9th, 2015, 08:34 PM   #4
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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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