document.write( "Question 187927: I am suppose to find the center (h,k) and the radius of the circle given by the equation: x^2-6x+y^2+10y=-30 by completing the square. \n" ); document.write( "

Algebra.Com's Answer #140892 by Earlsdon(6294) $\"\"$ $\"About$

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Find the center and the radius of the circle given by the equation:
\n" ); document.write( " $\"x%5E2-6x%2By%5E2%2B10y+=+-30\"$ First group the x- and y-variables as shown:
\n" ); document.write( " $\"%28x%5E2-6x%29%2B%28y%5E2%2B10y%29+=+-30\"$ Now complete the squares in both the x- and y-groups by adding the square of half the x-coefficient and the square of half the y-coefficient to both sides of the equation.
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Simplify this.
\n" ); document.write( " $\"%28x%5E2-6x%2B9%29%2B%28y%5E2%2B10y%2B25%29+=+-30%2B9%2B25\"$ Factor the x-trinomial and factor the y-trinomial and simplify the right side.
\n" ); document.write( " $\"%28x-3%29%5E2+%2B+%28y%2B5%29%5E2+=+4\"$ Now compare this with the standard form of the equation of a circle with its center at (h, k) and whose radius is r.
\n" ); document.write( " $\"%28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2\"$ and you can readily see that:
\n" ); document.write( " $\"h+=+3\"$
\n" ); document.write( " $\"k+=+-5\"$
\n" ); document.write( " $\"r+=+2\"$
\n" ); document.write( "So the center of the circle is at (3, -5) and the radius is 2. \n" ); document.write( "

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