document.write( "Question 160277: find the center and the radius of the circle x^2+2x+y^2-6y-6=0 \n" ); document.write( "
Algebra.Com's Answer #118231 by KnightOwlTutor(293)![]() ![]() You can put this solution on YOUR website! The first step you need to do us complete the sqaure\r \n" ); document.write( "\n" ); document.write( "Group the common variables together \r \n" ); document.write( "\n" ); document.write( "(x^2+2x+_)+(y^2-6y+_)=6\r \n" ); document.write( "\n" ); document.write( "Take the middle term and divide by 2 and get 1 square this number \n" ); document.write( "(x^2+2x+1)\r \n" ); document.write( "\n" ); document.write( "Do the same for the y variable\r \n" ); document.write( "\n" ); document.write( "y^2-6y+_)= -6/2=-3 (-3)^2=9\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "so now the equation looks like \n" ); document.write( "(x^2+2x+1)+(y^2-6y+9)= \r \n" ); document.write( "\n" ); document.write( "Now you have to add 1+9 to the other side of the equation to keep it the same \n" ); document.write( "(x^2+2x+1)+(y^2-6y+9)=6+1+9\r \n" ); document.write( "\n" ); document.write( "Factor both variables\r \n" ); document.write( "\n" ); document.write( "(x+1)^2+(y-3)^2=16\r \n" ); document.write( "\n" ); document.write( "the radius is at (-1,3) Radius is the sqrt of 16 and =4\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |