Question 1106860
One diameter is two radii.


{{{(5-(-3))^2+(6-(-4))^2}}}
{{{8^2+10^2}}}
{{{164}}}------the diameter squared
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{{{(2r)^2=164}}}


{{{4r^2=164}}}
{{{r^2=164/4}}}
{{{r^2=41}}}


Center of circle is midpoint of the two given diameter endpoints.
{{{x=(5-3)/2=1}}}
and
{{{(6-4)/2=2/2=1}}}
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center  (1,1).


EQUATION OF CIRCLE:  {{{highlight((x-1)^2+(y-1)^2=41)}}}