document.write( "Question 139392This question is from textbook Algebra 2
\n" ); document.write( ": What is the answer in standard form and the center and radius of this circle equation?
\n" ); document.write( "3x^2+3y^2-18x+6y=18
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Algebra.Com's Answer #101648 by solver91311(24713)\"\" \"About 
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You mean the 'equation' in standard form, I think.\r
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\n" ); document.write( "\n" ); document.write( "The standard form of an equation of a circle with center at (h,k) and radius r is \"%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "Notice that the terms on the left are each a squared binomial, so the trick to converting an equation in the form \"Ax%5E2%2BBy%5E2%2BCx%2BDy%2BC=0\" into standard form is to complete the square on each of the variables.\r
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\n" ); document.write( "\n" ); document.write( "\"3x%5E2%2B3y%5E2-18x%2B6y=18\"\r
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\n" ); document.write( "\n" ); document.write( "Notice that all of the coefficients on this one are divisible by 3, so divide through by 3 first because we need the coefficients on the 2nd degree terms to be 1.\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2%2By%5E2-6x%2B2y=6\" is an equivalent equation and describes the same conic section.\r
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\n" ); document.write( "\n" ); document.write( "Next rearrange the terms so that the x terms are together and the y terms are together:\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2-6x%2By%5E2%2B2y=6\"\r
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\n" ); document.write( "\n" ); document.write( "Divide the coefficient on the 1st degree x term by 2, square the result, and add that result to both sides of the equation: \"%28-6%2F2%29%5E2=%28-3%29%5E2=9\"\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2-6x%2B9%2By%5E2%2B2y=6%2B9\"\r
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\n" ); document.write( "\n" ); document.write( "Do the same thing with the coefficient on the 1st degree y term:
\n" ); document.write( "\"%282%2F2%29%5E2=1%5E2=1\", and collect terms on the right\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E2-6x%2B9%2By%5E2%2B2y%2B1=6%2B9%2B1\"
\n" ); document.write( "\"x%5E2-6x%2B9%2By%5E2%2B2y%2B1=16\"\r
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\n" ); document.write( "\n" ); document.write( "Now we have a sum of two perfect square trinomials: \"x%5E2-6x%2B9=%28x-3%29%5E2\" and \"y%5E2%2B2y%2B1=%28y%2B1%29%5E2\", so:\r
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\n" ); document.write( "\n" ); document.write( "\"%28x-3%29%5E2%2B%28y%2B1%29%5E2=16\", but \"16=4%5E2\" so:\r
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\n" ); document.write( "\n" ); document.write( "\"%28x-3%29%5E2%2B%28y%2B1%29%5E2=4%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "Now compare this result to the standard form \"%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2\". We need to make one little adjustment to make our result look exactly like the standard form pattern:\r
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\n" ); document.write( "\n" ); document.write( "\"%28x-3%29%5E2%2B%28y-%28-1%29%29%5E2=4%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "Now we can read the coordinates of the center and the value of the radius directly.\r
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\n" ); document.write( "\n" ); document.write( "\"h=3\", \"k=-1\", so the center is at (3,-1)\r
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\n" ); document.write( "\n" ); document.write( "and the radius is \"r=4\"\r
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