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finding the centre and radius |
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| Mar18-08, 08:36 PM | #1 |
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finding the centre and radius
The question states:
Find the centre and radius of each circle with equations as given (a) 3x^2 + 3y^2 = 81 (b) x^2 = 6y - y^2 I really don't know how to approach this question, i started (a) by dividing both sides by 3 but then i don't know where to go from there, and i don't even know how to do (b)...please help me
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| Mar18-08, 08:43 PM | #2 |
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Mentor
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Well, you know that a circle with centre (p,q) and radius r can be written (x-p)2+(y-q)2=r2.
(a) once you've divided through by 3, it is in the above form, isn't it? (b) try to put it in the above form by, maybe, completing the square for the y component. |
| Mar18-08, 09:04 PM | #3 |
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ok but cristo, for (a) you're saying that my centre would just be p and q?..no actual numbers?...and for (b) my radius would be 9? -also for (b) what would be my centre after completing the square for the y component?
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| Mar18-08, 09:14 PM | #4 |
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finding the centre and radius
a) P and Q are dummy variables. In your first problem: 3x^2 + 3y^2 = 81 you could think of this as 3(x-0)^2 + 3(y-0)^2 = 81
b) Remember the equation is (x-p)^2+(y-q)^2=r^2, make sure you're accounting for the radius^2. The center is straight forward once you have completed the square. |
| Mar18-08, 09:50 PM | #5 |
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...thank u guys
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