document.write( "Question 1143791: find the area enclosed by the curve x^2+y^2-3x-8y+18=0 \n" ); document.write( "

Algebra.Com's Answer #764696 by ikleyn(52787) $\"\"$ $\"About$

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document.write( "The curve is a circle, and all you need is to find its radius.\r\n" );
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document.write( "For it, transform the given equation of a circle from its general form to its standard form.\r\n" );
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document.write( "For it, complete the squares for x- and y-terms separately in the given equation\r\n" );
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document.write( "    x^2 + y^2 - 3x - 8y + 18 = 0\r\n" );
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document.write( "    (x^2 - 3x) + (y^2 - 8y) = -18\r\n" );
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document.write( "    (x^2 - 3x + 2.25) + (y^2 - 8y + 16) = -18 + 2.25 + 16\r\n" );
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document.write( "     +  = 0.25.\r\n" );
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document.write( "Thus you have the circle centered at the point (1.5,4) and having the radius of  r =  = 0.5.\r\n" );
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document.write( "The area of this circle is   =  = .    ANSWER\r\n" );
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