Cartesian coordinate problem

1. Nov 6, 2007

tony873004

Identify the curve by finding a Cartesian equation for the curve.
r=2

My attempt:
r=2 makes a circle with a radius of 2, so:
$$\begin{array}{l} x^2 + y^2 = r^2 \\ y^2 = r^2 - x^2 \\ \\ y = \pm \sqrt {r^2 - x^2 } \\ \\ y = \pm \sqrt {2^2 - x^2 } \\ \\ y = \pm \sqrt {4 - x^2 } \\ \end{array}$$

But the back of the book simply says
??That's not an equation. It's a description. Doesn't an equation need to have an equal sign?

Last edited: Nov 6, 2007
2. Nov 6, 2007

Avodyne

Bad book. Lots of 'em out there.

3. Nov 6, 2007

mjsd

hehehehe.... I am sure your tutor is not going to deduct marks for saying that instead of x^2+y^2=2^2,.. that answer actually shows that you know what x^2+y^2=2^2 really means.

4. Nov 6, 2007

HallsofIvy

Staff Emeritus
Was there a reason for solving for y?
$y= \pm \sqrt{4- x^2}$ says nothing that $x^2+ y^2= 4$ doesn't and I would consider the second form simpler.

Relevant to your actual question, the problem did NOT say "find the equation"- it said "Identify the curve by finding a Cartesian equation for the curve."

5. Nov 6, 2007

tony873004

But it told me how to identify the curve: Identify the curve by finding a Cartesian equation.

The teacher agreed with you. $x^2+ y^2= 4$ is a better way to state the answer. She said equations usually don't have a +/- in them.