Question

a circle is defined by the equation x^2 + y^2 - x - 2 y - 11/4 = 0. what are the coordinates for the center of the circle and the length of the radius?

Answer 1

please help

Answer 2

you need to complete the square on both x and y to put it in the form \[(x-h)^2 + (y-k)^2 =r^2\]from which you can read off the center \((h,k)\) and the radius \(r\)

do you know how to do that?

Answer 3

no, i never learned this

Answer 4

you don't know how to complete the square at all?

Answer 5

that's okay, I just want to know where to start with the explanation...

Answer 6

completing the square means we take something that looks like \[ax^2+bx+c\](where \(a,b,c\) are arbitrary values, and might even be 0) and turn it into something that looks like \[(x+a)^2\]

if we multiply out \[(x+a)^2 = (x+a)(x+a) = x(x+a) + a(x+a) = x^2 + ax + ax + a^2\]\[\qquad = x^2+2ax+a^2\]

That means if we have the first two terms (the \(x^2\) term and the \(x\) term), we take the coefficient of the \(x\) term, set that equal to \(2a\), and add \(a^2\). Our perfect square is then \((x+a)^2\)

we can see that if we have something like \[x^2 + 6x\]for example, we can find the number we need to add on to it to make it a perfect square by taking the coefficient of the non-squared term (which is 6 in the example), dividing it in half, and squaring that number. Our "perfect square" will be \[(x+\frac{6}{2})^2 = (x+3)^2\]because \[(x+3)(x+3) = x^2+3x+3x+3*3 = x^2+6x+9\]

Now, we need to do something to account for that 9. We can either subtract it from an existing number in the equation, or we can add it to both sides of the equation, or both add and subtract it on the same side of the equation.

Let's do a similar problem to yours as an example of the whole thing:

\[x^2+y^2 +2x + 4y = 0\]
I'm going to rewrite it with parentheses:
\[(x^2+2x) + (y^2+4y) = 0\]Now I complete the square on the \(x\) part:
\[(x^2+2x + (\frac{2}{2})^2) + (y^2 + 4y) = 0 + (\frac{2}{2})^2\]\[(x^2+2x + 1) + (y^2 + 4y) = 0 + 1\]\[(x+1)^2 + (y^2+4y) = 1\]Now I complete the square on the \(y\) part:
\[(x+1)^2 + (y^2 + 4y + (\frac{4}{2})^2) = 1 + (\frac{4}{2})^2\]\[(x+1)^2 + (y^2+4y+4) = 1+4\]\[(x+1)^2+(y+2)^2 = 5\]
Comparing that with the form of the equation for a circle:
\[(x-h)^2 + (y-k)^2 = r^2\]\[+1 = -h\]\[h=-1\]\[+2=-k\]\[k=-2\]\[r^2=5\]\[r=\sqrt{5}\]The center of the circle in my example is at \((h,k) = (-1,-2)\) and the radius is \(r = \sqrt{5}\)

That should give you everything you need to do your problem...