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**RauLiTo****Member**- Registered: 2006-01-11
- Posts: 142

get the equation of the circle which touch ( 0 , Y ) and ( - X , 0 ) r = 3 and

notice : the circle in the second quarter

sorry my english is not that much so i'll insert a picture :

sorry again for this bad english !

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

That's OK, we can understand you perfectly. Especially with your nifty picture.

Anyway, the formula for a circle centred at the origin is x² + y² = r², where r is the radius.

To move the centre, we need to change what we square accordingly.

We need to add or subtract how much the centre would need to move in each direction to get to the origin.

Your circle is centred at (-3, 3), so to be centred at the origin, it would need to move 3 in the x direction and -3 in the y direction.

And your radius is 3, so you want the right hand side to be 9.

Therefore, the equation for your circle is (x+3)² + (y-3)² = 9.

Why did the vector cross the road?

It wanted to be normal.

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**RauLiTo****Member**- Registered: 2006-01-11
- Posts: 142

thank you very much ma friend ... thats really kind from you :D

ImPo$$!BLe = NoTH!nG

Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

while its not necessary here, you can also use derivatives to help you find the point of tangency of a line tangent to a circle (or any differentiable function). If you know the slope of the line its tangent to. Sometimes it can speed things up.

x^2 + y^2 = r^2

2x + 2y dy/dx = 0 (r is a constant)

2y dy/dx = -2x

dy/dx = -x/y

so if a line is tangent to a circle, the slope of the line will be equal to the slope of the circle at the point of tangency. So if the circle is tangent to a line with a slope of 1/2 then -x/y = 1/2 this gives you the equation -2x = y, so you can substitute for y in the equation x^2 + y^2 = r^2.

Note, a circle reaches a given slope at exactly two points, so you must check both points to see which falls into the equation of the tangent line. (basicly subsitute the x and y values into the linear equation and see if it comes out true).

*Last edited by mikau (2006-03-10 10:28:38)*

A logarithm is just a misspelled algorithm.

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