Find an expression for the function whose graph is the given curve.
The top half of the circle x^2 + (y-2)^2 = 4
Please Help. I don't even know where to start.
See:
http://www.jiskha.com/display.cgi?id=1284510821
To find an expression for the function whose graph represents the top half of the circle given by the equation x^2 + (y-2)^2 = 4, you can start by solving for y.
Step 1: Start with the equation of the circle: x^2 + (y-2)^2 = 4.
Step 2: Expand the equation: x^2 + (y^2 - 4y + 4) = 4.
Step 3: Simplify the equation: x^2 + y^2 - 4y + 4 = 4.
Step 4: Subtract 4 from both sides: x^2 + y^2 - 4y = 0.
Step 5: Rearrange the terms: y^2 - 4y + x^2 = 0.
Step 6: Complete the square by adding (4/2)^2 = 4 to both sides to obtain: y^2 - 4y + 4 + x^2 = 4.
Step 7: Simplify the equation: y^2 - 4y + 4 + x^2 = 4.
Step 8: Rewrite the left side as a perfect square: (y-2)^2 + x^2 = 4.
Step 9: Subtract x^2 from both sides: (y-2)^2 = 4 - x^2.
Step 10: Take the square root of both sides: y - 2 = ±√(4 - x^2).
Step 11: Add 2 to both sides: y = 2 ± √(4 - x^2).
Now you have the expression for the top half of the circle given by the equation. The graph of this function will represent the top half of the circle x^2 + (y-2)^2 = 4.