Question 800663
Using the given line, the center of this circle must be on some general point (x,x+2).


One radius is {{{sqrt((x-2)^2+(x+2-2)^2)}}} and another radius is {{{sqrt((x-5)^2+(x+2-3)^2)}}}.
Simplifying somewhat, these radius expressions are {{{sqrt((x-2)^2+(x)^2)}}} and {{{sqrt((x-5)^2+(x-1)^2)}}}


{{{sqrt((x^2-4x+4)+(x^2))=sqrt((x^2-10x+25)+(x^2-2x+1))}}}
.
{{{2x^2-4x+4=2x^2-12x+26}}}
{{{-4x+4=-12x+26}}}
{{{8x+4=26}}}
{{{8x=22}}}
{{{x=22/8}}}
{{{highlight(x=11/4)}}}, this is much of the way to the answer, but solution is not yet complete.  You can finish this....?
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(Use line or general center point to find the actual center.  Compute the actual radius length.  Put information into standard form equation for a circle).