Question 939844
<pre> 
This involves solving several systems of equations,
either by substitution or elimination:

{{{drawing(200,640/3,-2,13,-7,9,graph(200,640/3,-2,13,-7,9),
line(-11,-11,15,15),line(-12,6,14,-7), line(0,35,14,-21),
circle(6.5,.5,6.519202405),locate(7,7,"(7,7)"),
locate(10,-5,"(10.-5)"),locate(0,0,"(0,0)")



  )}}}


First, find the vertices of the triangle by solving
the three systems of equations by either substitution
or elimination.

{{{system(x-y=0, x+2y=0)}}}  {{{system(x-y=0,4x+y=35)}}}  {{{system(x+2y=0,4x+y=35)}}}

The solutions are (0,0), (7,7) and (10,-5)

Those are points on the circle so they will satisfy
the equation:

{{{x^2+y^2+Dx+Ey+F=0}}}

Substituting the three vertices gives this system:

{{{system(0^2+0^2+D(0)+E(0)+F=0,7^2+7^2+D(7)+E(7)+F=0,10^2+(-5)^2+D(10)+E(-5)+F=0)}}} 

which simplifies to:

{{{system(F=0,98+7D+7E+F=0,125+10D-5E+F=0)}}}

And since F=0 we only have two equations:

{{{system(98+7D+7E=0,125+10D-5E=0)}}}

Dividing the first equation thru by 7 and the 2nd one by 5,

{{{system(14+D+E=0,25+2D-E=0)}}}

or

{{{system(D+E=-14,2D-E=-25)}}}

solving that systen gives D=-13, E=-1

So the general equation 

{{{x^2+y^2+Dx+Ey+F=0}}} bcomes

{{{x^2+y^2-13x-y=0}}}

Edwin</pre>