Question 968104
<pre> 
First we solve the system of equations: 

{{{system(x+3y=20, x^2+y^2-6x-8y=0)}}} 

to find where they intersect: 

Solve the first equation for x 
  
x+3y = 20 
   x = 20-3y 

Substitute (20-3y) for x in the 2nd equation: 

{{{x^2+y^2-6x-8y=0}}} 

{{{(20-3y)^2+y^2-6(20-3y)-8y=0}}} 

{{{400-120y+9y^2+y^2-120+18y-8y=0}}} 

{{{10y^2-110y+280 = 0}}} 

Divide thru by 10: 

{{{y^2-11y+28 = 0}}} 

{{{(y-7)(y-4)=0}}} 

y-7 = 0;  y-4 = 0 
  y = 7;    y = 4 

Substituting in 

x = 20-3y     x = 20-3y 
x = 20-3(7)   x = 20-3(4) 
x = 20-21     x = 20-12 
x = -1        x = 8 

So they intersect at P(-1,7) and Q(8,4) 

{{{drawing(400,400,-6,14,-6,14, graph(400,400,-6,14,-6,14), 
circle(3,4,5),line(-13,11,14,2), 
locate(-4,7,"P(-1,7)"),locate(8.1,4.7,"Q(8,4)") 

  

 )}}} 

Now we need to find the equation of another circle that has its
center as the midpoint of PQ, the green circle below:

{{{drawing(400,400,-6,14,-6,14, graph(400,400,-6,14,-6,14), 
circle(3,4,5),line(-13,11,14,2), 
locate(-4,7,"P(-1,7)"),locate(8.1,4.7,"Q(8,4)"),
green(circle(7/2,11/2,3sqrt(10)/2)) 

  

 )}}} 

Midpoint of PQ:  {{{(matrix(1,3,   (8+(-1))/2,",",(4+7)/2))}}}{{{""=""}}}{{{(matrix(1,3,   7/2,",",11/2))}}} 

and has its diameter as the length of PQ and its radius as half that:

{{{PQ=sqrt((8-(-1))^2+(4-7)^2)=sqrt(9^2+3^2)=sqrt(81+9)=sqrt(90)=sqrt(9*10)=3sqrt(10)}}} 

Therefore its center is {{{(matrix(1,3,7/2,",",11/2))}}} and its
radius is one-half the diameter {{{3sqrt(10)/2}}}.

Its equation is

{{{(x-h)^2+(y-k)^2=r^2}}}

{{{(x-7/2)^2+(y-11/2)^2=(3sqrt(10)/2)^2}}}

{{{(x-7/2)^2+(y-11/2)^2=(9*10)/4}}}

{{{(x-7/2)^2+(y-11/2)^2=90/4}}}

{{{(x-7/2)^2+(y-11/2)^2=45/2}}}

Or you can multiply it out and rearrange to get this:

{{{x^2+y^2-7x-11y+20 = 0}}}

Edwin</pre>