What is the Cartesian equation for a circle with a radius of 2?

[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}<br /> x^2 + y^2 = r^2 \\ <br /> y^2 = r^2 - x^2 \\ <br /> \\ <br /> y = \pm \sqrt {r^2 - x^2 } \\ <br /> \\ <br /> y = \pm \sqrt {2^2 - x^2 } \\ <br /> \\ <br /> y = \pm \sqrt {4 - x^2 } \\ <br /> \end{array}<br />

But the back of the book simply says

Circle, center O, radius 2

??That's not an equation. It's a description. Doesn't an equation need to have an equal sign?

 

[Avodyne]

Bad book. Lots of 'em out there.

 

[mjsd]

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}<br /> x^2 + y^2 = r^2 \\ <br /> y^2 = r^2 - x^2 \\ <br /> \\ <br /> y = \pm \sqrt {r^2 - x^2 } \\ <br /> \\ <br /> y = \pm \sqrt {2^2 - x^2 } \\ <br /> \\ <br /> y = \pm \sqrt {4 - x^2 } \\ <br /> \end{array}<br />

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?

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.

 

[HallsofIvy]

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."

 

[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.