document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=256775>Question 256775</A>: <I><SPAN STYLE=\"word-break: break-all;\"> wHAT IS THE CENTER OF THE CIRCLE X RAISED TO THE 2ND POWER + Y RAISED TO THE SECOND POWER - 4X + 2Y -11 = 0<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #188827 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=188827\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If I did this correctly, then the center of the circle should be (x,y) = (2,-1).\r<BR>\n" );
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document.write( "you wind up with the equation of:\r<BR>\n" );
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document.write( "(x-2)^2 + (y+1)^2 = 16\r<BR>\n" );
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document.write( "to graph this equation, you need to solve for y.\r<BR>\n" );
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document.write( "subtract (x-2)^2 from both sides of the equation to get:\r<BR>\n" );
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document.write( "(y+1)^2 = 16 - (x-2)^2\r<BR>\n" );
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document.write( "take square root of both sides of this equation to get:\r<BR>\n" );
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document.write( "y+1 = +/- sqrt(11-(x-2)^2)\r<BR>\n" );
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document.write( "subtract 1 from both sides of this equation to get:\r<BR>\n" );
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document.write( "y = -1 +/- sqrt(16-(x-2)^2)\r<BR>\n" );
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document.write( "graph this equation to get:\r<BR>\n" );
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document.write( "from the graph you can see that the center of the circle appears to be x,y = 2,-1.\r<BR>\n" );
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document.write( "if you freeze y at -1, then the horizontal range is from x = -2 to x = 6 with x = 2 right in the middle (+/- 4 each way).\r<BR>\n" );
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document.write( "if you freeze x at 2, then the vertical range is from y = -5 to y = 3 with y = -1 right in the middle (+/- 4 each way).\r<BR>\n" );
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document.write( "the center of the circle is definitely at x,y = (2,-1).\r<BR>\n" );
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document.write( "here's how it was derived:\r<BR>\n" );
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document.write( "your original equation is:\r<BR>\n" );
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document.write( "x^2 + y^2 - 4x + 2y - 11 = 0\r<BR>\n" );
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document.write( "reorder the terms to that the x's and the y's are together to get:\r<BR>\n" );
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document.write( "x^2 - 4x + y^2 + 2y - 11 = 0\r<BR>\n" );
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document.write( "complete the squares on x^2 - 4x to get (x-2)^2 - 4\r<BR>\n" );
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document.write( "complete the squares on y^2 + 2y to get (y+1)^2 - 1\r<BR>\n" );
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document.write( "your equation becomes:\r<BR>\n" );
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document.write( "(x-2)^2 - 4 + (y+1)^2 - 1 - 11 = 0\r<BR>\n" );
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document.write( "combine like terms to get:\r<BR>\n" );
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document.write( "(x-2)^2 + (y+1)^2 - 16 = 0\r<BR>\n" );
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document.write( "add 16 to both sides of the equation to get:\r<BR>\n" );
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document.write( "(x-2)^2 + (y+1)^2 = 16\r<BR>\n" );
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document.write( "you have just converted to the standard form of the equation for the circle.\r<BR>\n" );
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document.write( "that standard form is:\r<BR>\n" );
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document.write( "(x-h)^2 + (y-k)^2 = r^2\r<BR>\n" );
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document.write( "h is the x coordinate of the center of the circle.<BR>\n" );
document.write( "k is the y coodinate of the center of the circle.<BR>\n" );
document.write( "r is the radius of the circle.\r<BR>\n" );
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document.write( "the answer to problem is:\r<BR>\n" );
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document.write( "the center of the circle is at (x,y) = (2,-1).\r<BR>\n" );
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