Question 1012535
<pre>
{{{drawing(400,400,-7,8,-7,8,

graph(400,400,-8,7,-7,8),green(line(-9,-5,9,-5),line(2,9,2,-9)),

circle(1,-1,0.15),circle(1,-1,0.13),circle(1,-1,0.11),circle(1,-1,0.09),circle(1,-1,0.07),circle(1,-1,0.05),circle(1,-1,0.03),circle(1,-1,0.01),

circle(1.25,1,0.15),circle(1.25,1,0.13),circle(1.25,1,0.11),circle(1.25,1,0.09),circle(1.25,1,0.07),circle(1.25,1,0.05),circle(1.25,1,0.03),circle(1.25,1,0.01),

locate(2.2,7.6,"B(2,b)"), locate(1.25,1,"(h,k)"), locate(1,-1,"(1,-1)"),
locate(1/2,-5.1,"A(a,-5)"),



circle(0.5,-5,0.15),circle(0.5,-5,0.13),circle(0.5,-5,0.11),circle(0.5,-5,0.09),circle(0.5,-5,0.07),circle(0.5,-5,0.05),circle(0.5,-5,0.03),circle(0.5,-5,0.01),

circle(2,7,0.15),circle(2,7,0.13),circle(2,7,0.11),circle(2,7,0.09),circle(2,7,0.07),circle(2,7,0.05),circle(2,7,0.03),circle(2,7,0.01),
line(1/2,-5,2,7), circle(1.25,1,sqrt(36.5625))   )}}}

Since (1,-1) is one-third of the way between A(a,-5) and B(2,b),

the distance from (1,-1) to B(2,b) is twice the distance from
A(a,-5) to (1,-1), so we use the distance formula to find this
equation:

   {{{sqrt((b-(-1))^2+(2-1)^2)}}}{{{""=""}}}{{{2*sqrt((1-a)^2+(-1-(-5))^2)}}} 

   {{{sqrt((b+1)^2+(1)^2)}}}{{{""=""}}}{{{2*sqrt((1-a)^2+(-1+5)^2)}}}

   {{{sqrt((b+1)^2+1)}}}{{{""=""}}}{{{2*sqrt((1-a)^2+4^2)}}}

   {{{sqrt((b+1)^2+1)}}}{{{""=""}}}{{{2*sqrt((1-a)^2+16)}}}

   {{{(b+1)^2+1)}}}{{{""=""}}}{{{4*((1-a)^2+16)}}}

   {{{b^2+2b+1+1)}}}{{{""=""}}}{{{4*(1-2a+a^2+16)}}}

   {{{b^2+2b+2)}}}{{{""=""}}}{{{4*(17-2a+a^2)}}}

   {{{b^2+2b+2)}}}{{{""=""}}}{{{68-8a+4a^2}}}

   {{{b^2+2b)}}}{{{""=""}}}{{{66-8a+4a^2}}}

We also know that the slope of the line from (1,-1) to B(2,b) is 
equal to the slope of the line from A(a,-5) to (1,-1), because
they are segments of the same line, so we use the slope formula
to get another equation in a and b:

{{{(b-(-1))/(2-1)}}}{{{""=""}}}{{{(-1-(-5))/(1-a)}}}

{{{(b+1)/(2-1)}}}{{{""=""}}}{{{(-1+5)/(1-a)}}}

{{{(b+1)/1}}}{{{""=""}}}{{{4/(1-a)}}}

{{{(1-a)(b+1)}}}{{{""=""}}}{{{4}}}

{{{b+1-ab-a}}}{{{""=""}}}{{{4}}}

{{{b-3}}}{{{""=""}}}{{{ab+a}}}

{{{b-3}}}{{{""=""}}}{{{a(b+1)}}}

{{{(b-3)/(b+1)}}}{{{""=""}}}{{{a}}}

Substitute in

{{{b^2+2b}}}{{{""=""}}}{{{66-8a+4a^2}}}

{{{b^2+2b}}}{{{""=""}}}{{{66-8((b-3)/(b+1))+4((b-3)/(b+1))^2}}}

{{{b^2+2b}}}{{{""=""}}}{{{66-(8(b-3))/(b+1)+4(b-3)^2/(b+1)^2}}}

Multiply through by the LCD = {{{(b+1)^2}}}

{{{(b^2+2b)(b+1)^2}}}{{{""=""}}}{{{66(b+1)^2-8(b-3)(b+1)+4(b-3)^2}}}

{{{(b^2+2b)(b^2+2b+1)}}}{{{""=""}}}{{{66(b+1)^2-8(b-3)(b+1)+4(b-3)^2}}}

{{{b^4+4b^3+5b^2+2b}}}{{{""=""}}}{{{66(b^2+2b+1)-8(b^2-2b-3)+4(b^2-6b+9)}}}

{{{b^4+4b^3+5b^2+2b}}}{{{""=""}}}{{{66b^2+132b+66-8b^2+16b+24+4b^2-24b+36}}}

{{{b^4+4b^3+5b^2+2b}}}{{{""=""}}}{{{62b^2+124b+126}}}

{{{b^4+4b^3-57b^2-122b-126}}}{{{""=""}}}{{{0}}}
 
One possible zero is -9

-9 | 1  4 -57 -122 -126
   |<u>   -9  45  108  126</u>
     1 -5 -12  -14    0

So we have factored the polynomial as

{{{(b+9)(b^3-5b^2-12b-14)}}}{{{""=""}}}{{{0}}}

A possible zero of the cubic is 7

7 | 1 -5 -12 -14
  |<u>    7  14  14</u>
    1  2   2   0

So we have further factored the polynomial as

{{{(b+9)(b-7)(b^2+2b+2)}}}{{{""=""}}}{{{0}}}

The real solutions are b=-9 and b=7

Substitute those in

{{{(b-3)/(b+1)}}}{{{""=""}}}{{{a}}}

{{{(-9-3)/(-9+1)}}}{{{""=""}}}{{{a}}}
{{{(-12)/(-8)}}}{{{""=""}}}{{{a}}}
{{{3/2}}}{{{""=""}}}{{{a}}}

{{{(7-3)/(7+1)}}}{{{""=""}}}{{{a}}}
{{{4/8}}}{{{""=""}}}{{{a}}}
{{{1/2}}}{{{""=""}}}{{{a}}}

So we have two possibilities to consider

A(1/2,-5) and B(2,7),  A(3/2,-5) and B(2,-9)  

We discard the second one because it is just a
point down below the line y=-5 which is twice
as far from (1,-1) as it is from (1,-1) to
(3/2,5).  But then (1,-1) would then not be a 
point of trisection of AB. So the only points 
to consider for endpoints of the diameter are 

A(1/2,-5) and B(2,7).  

The center of the desired circle is the 
midpoint of AB, which we find by the midpoint
formula:

(h,k) = {{{(matrix(1,3,(1/2+2)/2,",",(-5+7)/2))}}}

(h,k) = {{{(matrix(1,3,(5/2)/2,",",2/2))}}} 

(h,k) = {{{(matrix(1,3,5/4,",",1))}}}

To find the radius we use the distance formula
from that center to B(2,7) 

r = {{{sqrt((2-5/4)^2+(7-1)^2)}}}

r = {{{sqrt((8/4-5/4)^2+6^2)}}}

r = {{{sqrt((3/4)^2+36)}}}

r = {{{sqrt(9/16+36)}}}

r = {{{sqrt(9/16+576/16)}}}

r = {{{sqrt(585/16)}}}

r = {{{sqrt(585)/4}}}

r = {{{9sqrt(65)/4}}}

So the standard equation of the circle is  {{{(x-h)^2+(y-k)^2=r^2}}}

{{{(x-5/4)^2+(y-1)^2}}}{{{""=""}}}{{{585/16}}}

But the problem asks for the general form, so we must get that in
the form {{{x^2+y^2+Dx+Ey+F=0}}}

You can do that part.  Here's the answer:

{{{x^2+y^2-expr(5/2)x-2y-34}}}{{{""=""}}}{{{0}}}

Edwin</pre>