Question 968131
<pre>
We need to find the equation of the circle that has its
center as the midpoint between (2,-1) and (6,-5), the green circle below:

{{{drawing(400,400,-3,8,-7,4, graph(400,400,-3,8,-7,4), 
circle(4,-3,5.656854249/2),line(2,-1,6,-5),
locate(.8,-.6,"(2,-1)"),locate(6,-5,"(6,-5)"),
circle(4,-3,.1)  

  

 )}}} 

Midpoint between (2,-1) and (6,-5), {{{(matrix(1,3,   (2+6)/2,",",(-1+(-5))/2))}}}{{{""=""}}}{{{(matrix(1,3,   8/2,",",(-6)/2))}}} = (4,-3) 

and has its diameter as the distance between between (2,-1) and (6,-5)
and its radius as half that:

{{{diameter=sqrt((6-2)^2+(-5-(-1))^2)=sqrt(4^2+(-4)^2)=sqrt(16+16)=sqrt(32)=sqrt(16*2)=4sqrt(2)}}} 

Therefore its center is (4,-3) and its
radius is one-half the diameter {{{2sqrt(2)}}}.

Its equation is

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

{{{(x-4)^2+(y-(-3))^2=(2sqrt(2))^2}}}

{{{(x-4)^2+(y+3)^2=4*2}}}

{{{(x-4)^2+(y+3)^2=8}}}

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

{{{x^2+y^2-8x+6y+17 = 0}}}

Edwin</pre>