document.write( "Question 816642: Please can you help solve this problem for me. The point (x,y) lies on a cirle which has the segment from (0,-1) to (1/3, 4/3) as a diameter. Show this is algebraic equation \n" ); document.write( "
Algebra.Com's Answer #491791 by LinnW(1048)\"\" \"About 
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Here we are looking for a circle with its center at the midpoint between
\n" ); document.write( " (0,-1) and (1/3,4/3). The midpoint is ( (0+1/3)/2 , (-1+4/3)/2 ).
\n" ); document.write( " ( 1/6 , 1/6 ). We also need the length of the radius of the circle
\n" ); document.write( " which has as diameter (0,-1) and (1/3,4/3). The length of the diameter
\n" ); document.write( " is sqrt( (y2-y1)^2 + (x2-x1)^2 ).
\n" ); document.write( " So we want sqrt ( ( 1/3 - 0 )^2 + ( 4/3 - (-1) )^2 )
\n" ); document.write( " = sqrt ( ( 1/3 )^2 + ( 4/3 + 1 )^2 )
\n" ); document.write( " = sqrt ( 1/9 + (7/3)^2 )
\n" ); document.write( " = sqrt ( 1/9 + 49/9 )
\n" ); document.write( " = sqrt ( 50/9 )
\n" ); document.write( " The radius is half of this or 25/9 .\r
\n" ); document.write( "\n" ); document.write( " The standard form for a circle is (x-h)^2 + (y-k)^2 = r^2
\n" ); document.write( " For our application h= 1/6 and k= 1/6 , with r = 25/9 r^2 = 625/81
\n" ); document.write( " So (x- 1/6)^2 + (y- 1/6)^2 = 625/81\r
\n" ); document.write( "\n" ); document.write( " Double check this. Plotting the above equation
\n" ); document.write( " http://www.wolframalpha.com/input/?i=%28x-+1%2F6%29^2+%2B+%28y-+1%2F6%29^2+%3D+625%2F81 \n" ); document.write( "
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