document.write( "Question 137240: I need to know what the center and radius of the following circle would be:\r
\n" ); document.write( "\n" ); document.write( "x^2+y^2-4x+4y+4=0\r
\n" ); document.write( "\n" ); document.write( "anything else you might be able to tell me about the circle would be very helpful (ie. ecentricity)\r
\n" ); document.write( "\n" ); document.write( "Thank you \n" ); document.write( "

Algebra.Com's Answer #100414 by Earlsdon(6294) $\"\"$ $\"About$

You can put this solution on YOUR website!
Start with the general form of the equation of a circle with center at (h, k) and radius r.
\n" ); document.write( " $\"%28x-h%29%5E2%2B%28y-k%29%5E2=+r%5E2\"$ , so we need to get your equation looking like this:
\n" ); document.write( " $\"x%5E2%2By%5E2-4x%2B4y%2B4+=+0\"$ Group the x-terms and the y-terms.
\n" ); document.write( " $\"x%5E2-4x%2B4%29%2By%5E2%2B4y%29+=+0\"$ The constant 4 could just as well have been added to the y-terms rather than the x-terms, so it doesn't make no never mind (oooh!)
\n" ); document.write( "Now we need to complete the square in both the x-terms and the y-terms, but, you'll notice that the x-terms are already there, so all we need to do is to add, to the y-terms, the square of half the y-coefficient ( $\"%284%2F2%29%5E2+=+4\"$ ) to both sides of the equation.
\n" ); document.write( " $\"%28x%5E2-4x%2B4%29%2B%28y%5E4%2B4y%2B4%29+=+0%2B4\"$ Now we factor the x- and y-trinomials to get:
\n" ); document.write( " $\"%28x-2%29%5E2+%2B+%28y%2B2%29%5E2+=+4\"$ and compare this with the general form:
\n" ); document.write( " $\"%28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2\"$ and you can easily see that:
\n" ); document.write( " $\"h+=+2\"$ , $\"k+=+-2\"$ and $\"r%5E2+=+4\"$ , so...
\n" ); document.write( "The center is at (2, -2) and the radius is 2
\n" ); document.write( "That's about all we can say about the circle.
\n" ); document.write( "You ask about the eccentricity?
\n" ); document.write( "Well, remember that \"eccentricity\" is a measure of the \"flatness\" and applies to ellipses. Of course, a circle has no flatness, so the eccentricity is zero.
\n" ); document.write( "Algebraically, eccentricity is: $\"e+=+c%2Fa\"$ where c is the distance from one focus to the center (for a circle, this is zero) and a is the length of the semi-major axis (of an ellipse). \n" ); document.write( "

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