document.write( "Question 694253: find the center and radius of a circle (3x^2)+(2x)+(3y^2)-(7y)-(10)=0 I would like to see the steps as you go along to see how you find the answer. Thank you so much! \n" ); document.write( "
Algebra.Com's Answer #428167 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
DISCLAIMER: I do not believe in memorizing multi-step recipes (not my religion), so I will give the logic rationale for a way to solve the problem.
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\n" ); document.write( "The Pythagorean theorem and the drawing below show why \"%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2\" represents a circle with center (h,k) and radius \"r\" . (PREACHING: Memorizing \"%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2\" as a formula does not make sense).
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\n" ); document.write( "We have to find the coordinates of the center (h and k), and the radius r, and with very little extra work, we can get the equation in the form \"%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2\".
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\n" ); document.write( "From \"3x%5E2%2B2x%2B3y%5E2-7y-10=0\", you could to get to something that looks like
\n" ); document.write( "\"%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2\" where (h,k) represents the center of the circle and \"r\" is the radius.
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\n" ); document.write( "\"3x%5E2%2B2x%2B3y%5E2-7y-10=0\" --> \"%283x%5E2%2B2x%29%2B%283y%5E2-7y%29=10\" --> \"%283x%5E2%2B2x%29%2F3%2B%283y%5E2-7y%29%2F3=10%2F3\" --> \"%28x%5E2%2B2x%2F3%29%2B%28y%5E2-7y%2F3%29=10%2F3\"
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\n" ); document.write( "\"x%5E2%2B2x%2F3\" is not a perfect square. To make it into something like
\n" ); document.write( "\"%28x-h%29%5E2=x%5E2-2hx%2Bh%5E2\" we would have to add a third term.
\n" ); document.write( "(That is called \"completing the square\", a vocabulary term that often needs to be memorized, to speak the same language as the teachers. I do not object to memorizing a few terms, and I'll allow that one, because it's useful).
\n" ); document.write( "For all \"x\", \"-2hx=2x%2F3\" , so \"highlight%28h=-1%2F3%29\" gives us the x-coordinate of the center,
\n" ); document.write( "and \"h%5E2=1%2F9\" is the third term we have to add.
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\n" ); document.write( "The same can be done for \"%28y%5E2-7y%2F3%29\" to get it to look like \"%28y-k%29%5E2=y%5E2-2ky%2Bk%5E2\"
\n" ); document.write( "\"-2ky=-7y%2F3\" --> \"highlight%28k=7%2F6%29\", finding the y-coordinate of the center,
\n" ); document.write( "and finding \"k%5E2=49%2F36\" , the third term we have to add to complete the square.
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\n" ); document.write( "If we add those third terms to the left side of the equal sign we have to add them to the right side too, so
\n" ); document.write( "\"%28x%5E2%2B2x%2F3%29%2B%28y%5E2-7y%2F3%29=10%2F3\" --> \"%28x%5E2%2B2x%2F3%2B1%2F9%29%2B%28y%5E2-7y%2F3%2B49%2F36%29=10%2F3%2B1%2F9%2B49%2F36\"
\n" ); document.write( "Adding the fractions on the right side, and writing the left side as the sum of squares, we get the equation of the circle in the form we want:
\n" ); document.write( "\"%28x%5E2%2B2x%2F3%2B1%2F9%29%2B%28y%5E2-7y%2F3%2B49%2F36%29=10%2F3%2B1%2F9%2B49%2F36\" --> \"%28x-1%2F3%29%5E2%2B%28y-7%2F6%29%5E2=12%2F36%2B4%2F36%2B49%2F36\" --> \"%28x-1%2F3%29%5E2%2B%28y-7%2F6%29%5E2=120%2F36%2B4%2F36%2B49%2F36\" --> \"highlight%28%28x-1%2F3%29%5E2%2B%28y-7%2F6%29%5E2=173%2F36%29\"
\n" ); document.write( "(If we did not care about the equation, all we needed to do is add on the right side).
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\n" ); document.write( "Now we find the radius \"r%5E2=173%2F36\" --> \"highlight%28r=sqrt%28173%2F6%29%29\" \n" ); document.write( "
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