Question 826407
For each line, the gradient (we call it slope) is the differences in y-coordinates divided by the difference in x-coordinates.
 
Segment AB is part of the horizontal line {{{y=0}}} , the x-axis.
Its midpoint is (0,0) , the origin.
The perpendicular bisector to AB is the line perpendicular to the x-axis through the origin. That is the line {{{x=0}}} , the y-axis.
All that is so by design (clever choice of coordinates, placing the triangle so it would be so).
 
The other two perpendicular bisectors are not so easy to find.
 
The gradient/slope of AC is
{{{m[AC]=(y[C]-y[A])/(x[C]-x[A])=(2c-0)/(2b-2a)=2c/(2(b-a))=c/(b-a)}}} .
A perpendicular line would have a gradient/slope of
{{{(-1)/(c/(b-a))=-(b-a)/c=(a-b)/c}}}
The midpoint of AC has
{{{x=(x[C]+x[A])/2=(2b+2a)/2=2(b+a)/2=b+a}}}
{{{y=(y[C]+y[A])/2=(2c+0)/2=2c/2=c}}}
With that we can write the equation for the perpendicular bisector to AC in point-slope form as
{{{y-c=((a-b)/c)(x-(b+a))}}}
The intersection of that line and the line {{{x=0}}} has {{{x=0}}} (of course) and
{{{y-c=((a-b)/c)(0-(b+a))}}}
{{{y-c=-(a-b)(b+a)/c}}}
{{{y=(b^2-a^2)/c+c}}}
{{{y=(c^2+b^2-a^2)/c}}}
 
Similarly,
the midpoint of BC has
{{{x=(x[C]+x[B])/2=(2b-2a)/2=2(b-a)/2=b-a}}}
{{{y=(y[C]+y[B])/2=(2c+0)/2=2c/2=c}}}
The gradient of BC is
{{{m[BC]=(y[C]-y[B])/(x[C]-x[B])=(2c-0)/(2b+2a)=2c/(2(b+a))=c/(b+a)}}} .
A perpendicular line would have a gradient/slope of
{{{(-1)/(c/(b+a))=-(b+a)/c=-(a+b)/c}}}
With that we can write the equation for the perpendicular bisector to AC in point-slope form as
{{{y-c=(-(a+b)/c)(x-(b-a))}}}
So the intersection of that line and {{{x=0}}} has {{{x=0}}} (of course) and
{{{y-c=(-(a+b)/c)(0-(b-a))}}}
{{{y-c=(-(a+b)/c)(a-b)}}}
{{{y-c=-(a+b)(a-b)/c}}}
{{{y-c=-(a^2-b^2)/c}}}
{{{y-c=(b^2-a^2)/c}}}
{{{y=(b^2-a^2)/c+c}}}
{{{y=(c^2+b^2-a^2)/c}}}
 
So all 3 perpendicular bisectors pass through the point with
{{{x=0}}} and {{{y=(c^2+b^2-a^2)/c}}} .
Since all points on the perpendicular bisector of a segment are equidistant from the ends of the segment, the point above should be at the same distance from A, B, and C, so a circle centered at that point, with that equal distance as its radius would pass through A, B, and C.
Its radius can be calculated as the distance from that point to A(2a,0)
{{{sqrt((0-2a)^2+((c^2+b^2-a^2)/c-0)^2)=sqrt(4a^2+(c^2+b^2-a^2)^2/c^2)}}}