Find Center, Radius and Graph Each Circle

**symmetry** · January 15th 2007, 12:15 PM

Find the center (h, k) and radius r of each circle below.

Then, graph each circle.

(1) x^2 + (y - 1)^2 = 1

(2) x^2 + y^2 + x + y - (1/2) = 0

(3) How do I graph such circles WITHOUT a graphing calculator?

**topsquark** · January 15th 2007, 01:17 PM

Originally Posted by symmetry

Find the center (h, k) and radius r of each circle below.

Then, graph each circle.

(1) x^2 + (y - 1)^2 = 1

(2) x^2 + y^2 + x + y - (1/2) = 0

(3) How do I graph such circles WITHOUT a graphing calculator?

The general equation for a circle is $(x - h)^2 + (y - k)^2 = r^2$ where the center of the circle is (h, k) and the circle has a radius of r.

I won't even bother with the first one...

2) $x^2 + y^2 + x + y - \frac{1}{2} = 0$

We need to complete the square for both the x and y variables:
$(x^2 + x) + (y^2 + y) = \frac{1}{2}$

Now, $(x + a)^2 = x^2 + 2ax + a^2$ . So looking at the first set of parenthesis we've got $x^2 + x$ . Equating the linear coefficient of this expression and the one right above we see that $1 = 2a$ , so $a = \frac{1}{2}$ . This means we want to add $a^2 = \frac{1}{4}$ to $x^2 + x$ to make it a perfect square since $x^2 + x + \frac{1}{4} = \left ( x + \frac{1}{2} \right ) ^2$ . But what we do to one side of the equation we need to do to the other. Thus:
$\left ( x^2 + x + \frac{1}{4} \right ) + (y^2 + y) = \frac{1}{2} + \frac{1}{4}$

$\left ( x + \frac{1}{2} \right )^2 + (y^2 + y) = \frac{1}{2} + \frac{1}{4}$

Now to the y variable. The problem is identical and we find we need to add $\frac{1}{4}$ to both sides again:
$\left ( x + \frac{1}{2} \right )^2 + \left ( y^2 + y + \frac{1}{4} \right ) = \frac{1}{2} + \frac{1}{4} + \frac{1}{4}$

$\left ( x + \frac{1}{2} \right )^2 + \left ( y + \frac{1}{2} \right )^2 = 1$

So this circle has a center at $\left ( -\frac{1}{2}, -\frac{1}{2} \right )$ and a radius of 1.

-Dan

**topsquark** · January 15th 2007, 01:18 PM

Originally Posted by symmetry

(3) How do I graph such circles WITHOUT a graphing calculator?

Who said you had to use a graphing calculator to graph? Get some graph paper and work out some points by hand!

Hey, I had to do that in High School...

-Dan

**symmetry** · January 15th 2007, 01:30 PM

Dan,

I follow your steps except for one: letter a in the mix.

Where did you get the letter a?

I got lost when you put variable a into the steps.

What does a represent?

**topsquark** · January 15th 2007, 01:55 PM

Originally Posted by symmetry

Dan,

I follow your steps except for one: letter a in the mix.

Where did you get the letter a?

I got lost when you put variable a into the steps.

What does a represent?

a is just a "dummy variable." In order to complete the square for $x^2 + x$ we need to add some constant to it. So I am comparing
$x^2 + x$
$x^2 + 2ax + a^2$ <-- This is the form for the perfect square $(x + a)^2$ .

where $a^2$ is the number we need to add to $x^2 + x$ . By comparing the two expressions we can see that we need the linear x term to have the same coefficient in both lines, so $1 = 2a$ .

Hopefully that explains it well enough. If not I can provide other examples.

-Dan

**symmetry** · January 15th 2007, 03:59 PM

I got it now.

Thanks.

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