Question 623929
If you locate the center of the line
between (-9,-2) and ( 1,1 ), that will be the
center of the circle. The standard form for
the circle is {{{ (x-h)^2 + (y-k)^2 = r^2 }}} and 
the center is at ( h,k )
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The x-coordinate of the center is
{{{ ( 1 - 9 ) / 2 = -4 }}}
The y-coordinate of the center is
{{{ ( 1 -2 ) / 2 = - 1/2 }}}
{{{ h = -4 }}}
{{{ k = -1/2 }}}
The radius is 1/2 of the diameter
{{{ d = sqrt( ( 1 -(-9) )^2 + ( 1 -(-2) )^2 ) }}}
{{{ d = sqrt( 10^2 + 3^2 ) }}}
{{{ d = sqrt( 109 ) }}}
{{{ r =  sqrt(109)/ 2 }}}
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{{{ (x-h)^2 + (y-k)^2 = r^2 }}}
{{{ (x-(-4))^2 + (y-(-1/2))^2 = ( sqrt(109)/2)^2 }}}
{{{ ( x + 4 )^2 + ( y + 1/2 )^2 = 109/4 }}}
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check:
Does it pass through ( -9,-2 ) ?
{{{ ( -9 + 4 )^2 + ( -2 + 1/2 )^2 = 109/4 }}}
{{{ ( -5 )^2 + ( -3/2)^2 = 109/4 }}}
{{{ 25 + 9/4 = 109/4 }}}
{{{ 100/4 + 9/4 = 109/4 }}}
{{{ 109/4 = 109/4 }}}
OK- you can check (1,1)