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April 16, 2014

April 16, 2014

Posted by **Chelsea** on Tuesday, September 14, 2010 at 8:33pm.

The top half of the circle x^2 + (y-2)^2 = 4

Please Help. I don't even know where to start.

- Calc -
**MathMate**, Tuesday, September 14, 2010 at 8:44pmHints:

1. Solve for y in terms of x.

2. You will have an expression involving a square-root of an expression.

3. You will need to restrict the expression under the radical to be non-negative, or else you will end up with a complex quantity.

4. For the top half of the curve (circle), restrict the value of the square-root to be non-negative, i.e. take the positive value of the square-root only.

- Calc -
**Chelsea**, Tuesday, September 14, 2010 at 8:58pmSo what do you mean by y in terms of x? I solved x^2 + (y-2)^2 = 4 and I got y^2 - 4y + 8 = x^2 ?

- Calc -
**MathMate**, Tuesday, September 14, 2010 at 10:09pmI would isolate y on the left like:

(y-2)²=4-x²

and take square-root on both sides.

transpose the -2 to the right and you'll get your f(x)=y=....

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