document.write( "Question 1043573: For the circle whose equation is x-squir + y-squir -5y-14 equal to 0.
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Algebra.Com's Answer #658730 by ikleyn(52794)\"\" \"About 
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\n" ); document.write( "For the circle whose equation is x-squir + y-squir -5y-14 equal to 0.
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document.write( "You are given an equation of a curve\r\n" );
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document.write( "\"x%5E2+%2B+y%5E2+-+5y+-+14\" = \"0\".\r\n" );
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document.write( "Complete the squares separately for \"x\" and \"for \"y\".\r\n" );
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document.write( "\"x%5E2+%2B+%28y-2.5%29%5E2+-+6.25+-+14\" = \"0\"   ( I added and distracted the number 6.25:  \"y%5E2+-+5y\" = \"%28y%5E2+-+5y+%2B+6.25%29+-+6.25\" = \"%28y-2.5%5E2%29+-+6.25\" )\r\n" );
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document.write( "\"x%5E2+%2B+%28y-2.5%29%5E2+-+20.25\" = \"0\",\r\n" );
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document.write( "\"x%5E2+%2B+%28y-2.5%29%5E2\" = \"20.25\".\r\n" );
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document.write( "It is the equation of the circle with the center at the point (x,y) = (0, 2.5)  and the radius of \"sqrt%2820.25%29\" = 4.5.\r\n" );
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document.write( "Answer.  the circle with the center at the point (x,y) = (0, 2.5)  and the radius of 4.5.\r\n" );
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