Question 707202

<pre>Here's another way to do the problem:


{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),
locate(2,1,"(2,1)"), locate(3,5,"(3,5)"),
circle(2,1,.1), circle(3,5,.1), line(-19,32,16,-24) )}}}

Suppose (x,y) is the center of the circle.  Then the distance from
(x,y) to (2,1) must equal the distance from (x,y) to (3,5), so
we use the diatance formula and set them equal:

{{{d = sqrt( (x[2]-x[1])^2 + (y[2]-y[1])^2)}}}

{{{sqrt( (x-2)^2 + (y-1)^2)}}} = sqrt( (x-3)^2 + (y-5)^2)}}}

(x-2)² + (y-1)² = (x-3)² + (y-5)²

Rearrange to get difference of squares on each side:

(x-2)² - (x-3)² = (y-5)² - (y-1)²

[(x-2)-(x-3)][(x-2)+(x-3)] = [(y-5)-(y-1)][(y-5)+(y-1)]

[x-2-x+3]{x-2+x-3] = [y-5-y+1][y-5+y-1]

[1][2x-5] = [-4][2y-6]

2x-5 = -8y + 24

2x + 8y = 29

So we have the system of equations:

{{{system(2x + 8y = 29,8x + 5y = 8)}}}

Solve that and get the center, and do the rest like the
other solution.

Edwin</pre>