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 $\displaystyle (x - h)^2 + (y - k)^2 = r^2$ 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) $\displaystyle x^2 + y^2 + x + y - \frac{1}{2} = 0$
We need to complete the square for both the x and y variables:
$\displaystyle (x^2 + x) + (y^2 + y) = \frac{1}{2}$
Now, $\displaystyle (x + a)^2 = x^2 + 2ax + a^2$. So looking at the first set of parenthesis we've got $\displaystyle x^2 + x$. Equating the linear coefficient of this expression and the one right above we see that $\displaystyle 1 = 2a$, so $\displaystyle a = \frac{1}{2}$. This means we want to add $\displaystyle a^2 = \frac{1}{4}$ to $\displaystyle x^2 + x$ to make it a perfect square since $\displaystyle x^2 + x + \frac{1}{4} = \left ( x + \frac{1}{2} \right ) ^2$. But what we do to one side of the equation we need to do to the other. Thus:
$\displaystyle \left ( x^2 + x + \frac{1}{4} \right ) + (y^2 + y) = \frac{1}{2} + \frac{1}{4}$
$\displaystyle \left ( x + \frac{1}{2} \right )^2 + (y^2 + y) = \frac{1}{2} + \frac{1}{4}$
Now to the y variable. The problem is identical and we find we need to add $\displaystyle \frac{1}{4}$ to both sides again:
$\displaystyle \left ( x + \frac{1}{2} \right )^2 + \left ( y^2 + y + \frac{1}{4} \right ) = \frac{1}{2} + \frac{1}{4} + \frac{1}{4}$
$\displaystyle \left ( x + \frac{1}{2} \right )^2 + \left ( y + \frac{1}{2} \right )^2 = 1$
So this circle has a center at $\displaystyle \left ( -\frac{1}{2}, -\frac{1}{2} \right )$ and a radius of 1.
-Dan
a is just a "dummy variable." In order to complete the square for $\displaystyle x^2 + x$ we need to add some constant to it. So I am comparing
$\displaystyle x^2 + x$
$\displaystyle x^2 + 2ax + a^2$ <-- This is the form for the perfect square $\displaystyle (x + a)^2$.
where $\displaystyle a^2$ is the number we need to add to $\displaystyle x^2 + x$. By comparing the two expressions we can see that we need the linear x term to have the same coefficient in both lines, so $\displaystyle 1 = 2a$.
Hopefully that explains it well enough. If not I can provide other examples.
-Dan