# Curves

• Feb 10th 2009, 01:20 PM
s7b
Curves
I'm not sure what the curve of x=-(sqrt(1-(y-1)^2)) would look like....anyone know??
• Feb 10th 2009, 01:53 PM
Jameson
Quote:

Originally Posted by s7b
I'm not sure what the curve of x=-(sqrt(1-(y-1)^2)) would look like....anyone know??

Square both sides and add (y-1)^2 to both sides and you should see a familiar looking form.
• Feb 10th 2009, 01:57 PM
s7b
oh ok, so i think it would be a circle of radius one centered about (0,1) ??
• Feb 10th 2009, 02:03 PM
Jameson
Quote:

Originally Posted by s7b
oh ok, so i think it would be a circle of radius one centered about (0,1) ??

Sounds good. But you weren't given the ending formula, you were given the negative square root part. How does that affect the graph?
• Feb 10th 2009, 02:08 PM
s7b
Would that mean that it is reflected in the x-axis giving a circle of radius one centered at (0,-1)
• Feb 10th 2009, 02:13 PM
Jameson
When you take conics and apply a square root, the result is a split in the graph. In your problem x is always going to be negative so I believe this implies that it is the circle you first mentioned, cut in half vertically, and the left side of the half circle.