document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=283833>Question 283833</A>: <I><SPAN STYLE=\"word-break: break-all;\"> given the equation of a circle x^2+y^2-10x+4y+13=0. find its center and radius. i am having trouble please help </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #205977 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=205977\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x^2 + y^2 - 10x + 4y + 13 = 0\r<BR>\n" );
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document.write( "This equations is not in standard form.\r<BR>\n" );
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document.write( "Standard form of the equation of a circle is:\r<BR>\n" );
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document.write( "(x-h)^2 + (y-k)^2 = r^2 where:\r<BR>\n" );
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document.write( "(h,k) are the coordinates of the center of the circle and r is the radius of the circle.\r<BR>\n" );
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document.write( "You need to convert your equation into standard form in order to solve this problem.\r<BR>\n" );
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document.write( "Move all the terms around until all the x terms are together and all the y terms are together, and the constant term is on the right side of the equation.\r<BR>\n" );
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document.write( "You get:\r<BR>\n" );
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document.write( "x^2 - 10x + y^2 + 4y = -13\r<BR>\n" );
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document.write( "Complete the squares for the x terms and y terms.\r<BR>\n" );
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document.write( "x^2 - 10x = (x-5)^2 - 25\r<BR>\n" );
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document.write( "y^2 + 4y = (y+2)^2 - 4\r<BR>\n" );
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document.write( "Your equation becomes:\r<BR>\n" );
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document.write( "(x-5)^2 - 25 + (y+2)^2 - 4 = -13\r<BR>\n" );
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document.write( "Add 25 and add 4 to both sides of this equation to get:\r<BR>\n" );
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document.write( "(x-5)^2 + (y+2)^2 = -13 + 29 which becomes:\r<BR>\n" );
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document.write( "(x-5)^2 + (y+2)^2 = 16\r<BR>\n" );
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document.write( "The center of your circle should be (x,y) = (5,-2)\r<BR>\n" );
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document.write( "The radius of your circle should be sqrt(16) = 4\r<BR>\n" );
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document.write( "Graph your circle to see if this is true.\r<BR>\n" );
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document.write( "To graph your circle, you need to solve for y.\r<BR>\n" );
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document.write( "Your equation to work with is:\r<BR>\n" );
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document.write( "(x-5)^2 + (y+2)^2 = 16\r<BR>\n" );
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document.write( "Subtract (x-5)^2 from both sides of this equation to get:\r<BR>\n" );
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document.write( "(y+2)^2 = -(x-5)^2 + 16\r<BR>\n" );
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document.write( "Take the square root of both sides of this equation to get:\r<BR>\n" );
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document.write( "y+2 = +/- sqrt(-(x-5)^2 + 16)\r<BR>\n" );
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document.write( "Subtract 2 from both sides of this equation to get\r<BR>\n" );
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document.write( "y = +/- sqrt(-(x-5)^2 + 16) - 2\r<BR>\n" );
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document.write( "You get 2 equations out of this.\r<BR>\n" );
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document.write( "They are:\r<BR>\n" );
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document.write( "y = + sqrt(-(x-5)^2 + 16) - 2<BR>\n" );
document.write( "y = - sqrt(-(x-5)^2 + 16) - 2\r<BR>\n" );
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document.write( "Graph these 2 equations and you should have your circle.\r<BR>\n" );
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document.write( "The graph of these 2 equations is shown below:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28400%2C400%2C-10%2C10%2C-10%2C10%2Csqrt%28-%28x-5%29%5E2+%2B+16%29+-+2%2C-sqrt%28-%28x-5%29%5E2+%2B+16%29+-+2%2C-2%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "I drew a horizontal line at y = -2.\r<BR>\n" );
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document.write( "Draw an imaginary vertical line at x = 5 and you'll see that the center of the circle is at (x,y) = (5,-2).\r<BR>\n" );
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document.write( "When y = -2, the edge of the circle is at x = 1, and at x = 9.   Both of these are 4 units away from x = 5, proving that the radius of the circle is 4 units long.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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