document.write( "Question 1085764: Question: Identify the radius and center, if the graph of the equation is a circle then sketch it. Otherwise, specify if the graph is a single point (and what this point is), or has no point at all.\r
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document.write( "1.8x²+8y²-28+12y+221=0
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document.write( "2.x²+y²-8x+6y+40=0
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document.write( "3.x²+y²+22x+18y+215=0
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document.write( "4.x²+y²-18x+12y+144=0 \n" );
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Algebra.Com's Answer #699853 by Fombitz(32388)![]() ![]() You can put this solution on YOUR website! Complete the square to get the equation into the general equation of a circle. \n" ); document.write( " \n" ); document.write( "I'll do one, you do the others in the same way. \n" ); document.write( ". \n" ); document.write( ".\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "I'll assume you made a mistake and that it's \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "Since the radius term is negative, there is no point at all. \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( "If it equals zero, then the center is the single point. \n" ); document.write( "If its positive, then the square root of the value is the radius of the circle. \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( "Do the others the same way. \n" ); document.write( " \n" ); document.write( " |