Question 980038
Find an equation of the line containing the centers of the two circles: 
x² + y² - 10x - 10y + 49 = 0 and x² + y² - 4x - 6y + 9 = 0.
    
a. -2x - 3y + 5 = 0
    
b. 2x + 3y + 5 = 0
    
c. 8x - 7y + 5 = 0
    
d. 2x - 3y + 5 = 0

<pre>
I'll show you two ways, an easy way and a harder way,
though neither is hard:

Here's the easiest way:

 x² + y² - 10x - 10y + 49 = 0
 x² + y² -  4x -  6y +  9 = 0
-----------------------------
           -6x -  4y + 40 = 0 
or
            3x +  2y - 20 = 0

Subtracting the two equations of a circle gives the 
equation of the radical axis of the two circles, which 
is perpendicular to the line through their centers.

The slope of any line whose equation is Ax+By+C=0 has 
slope -A/B, so the radical axis has slope -3/2 and a
line perpendicular to it has slope 2/3.  As it turns
out, only choice d has slope 2/3.  So the answer is d.

However I think your instructor would not accept this
method although it is quite correct.  

Reason 1: There may have been more than one choice of 
equations of lines with slope 2/3, and you would not
be able to determine which one it was.

Reason 2:  You may not have studied the radical axis 
of two circles. 

----------------------------

The other way.

We find the centers of the two circles and find the equation 
of the line through them.

But there is an easy way to find the center of a circle
whose equation is

x² + y² + Dx + Ey + F = 0

The center is simply (h,k) = {{{(matrix(1,3,-D/2,",",-E/2))}}}

Therefore

 x² + y² - 10x - 10y + 49 = 0 has center (5,5)
 x² + y² -  4x -  6y +  9 = 0 has center (2,3)

So the slope is {{{(3-5)/(2-5)=(-2)/(-3)=2/3}}}

And equation:  y-3 = {{{2/3}}}(x-2)
              3y-9 = 2(x-2)
              3y-9 = 2x-4
          -2x+3y-5 = 0
           2x-3y+5 = 0

Choice d. 

Edwin</pre>