SOLUTION: 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 +

Algebra ->  Graphs -> SOLUTION: 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 +      Log On


   

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
Found 3 solutions by Cromlix, MathLover1, Edwin McCravy:
Answer by Cromlix(4381) About Me  (Show Source):
You can put this solution on YOUR website!
Hi there,
x2 + y2 - 10x - 10y + 49 = 0 and x2 + y2 - 4x - 6y + 9 = 0
First circle has a centre (5,5)
Second circle has a centre (2,3)
Gradient of line
y2 - y1/x2 - x1
3 - 5/2 - 5
-2/-3
= 2/3
Using line equation:
y - b = m(x - a) and point (5,5)
y - 5 = 2/3(x - 5)
y - 5 = 2/3x - 10/3
y = 2/3x - 10/3 + 15/3 (5)
y = 2/3x + 5/3
Multiply through by 3
3y = 2x + 5
2x - 3y + 5 = 0
Hope this helps:-)

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
an equation of the line containing the centers of the two circles:
x%5E2+%2B+y%5E2+-+10x+-+10y+%2B+49+=+0 ...........complete squares and write equation of a circle in a form %28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2
%28x%5E2+-+10x%2Bb%5E2%29+-b%5E2%2B+%28y%5E2+-+10y%2Bb%5E2%29+-b%5E2%2B+49+=+0
....recall, %28a-b%29%5E2=a%5E2-2ab%2Bb%5E2
in your case %28x%5E2+-+10x%2Bb%5E2%29, a=1 and 2ab=10, so we have 2%2A1%2Ab=10=>2b=10=>b=5
and we can write
%28x%5E2+-+10x%2B5%5E2%29+-5%5E2%2B+%28y%5E2+-+10y%2B5%5E2%29+-5%5E2%2B+49+=+0
%28x+-+5%29%5E2+-25%2B+%28y+-+5%29%5E2+-25%2B+49+=+0
%28x+-+5%29%5E2+%2B+%28y+-+5%29%5E2+-50%2B+49+=+0
%28x+-+5%29%5E2+%2B+%28y+-+5%29%5E2+-1+=+0
%28x+-+5%29%5E2+%2B+%28y+-+5%29%5E2++=+1
as you can see, h=5,k=5, and r=1
so, the center is at (5,5)

now do same with other circle:
x%5E2+%2B+y%5E2+-+4x+-+6y+%2B+9+=+0
%28x%5E2++-+4x%2Bb%5E2%29-b%5E2%2B+%28y%5E2+-+6y+%2Bb%5E2%29-b%5E2%2B+9+=+0......2ab=4=>2%2A1%2Ab=4=>b=2 and for y we have 2ab=6=>2%2A1%2Ab=6=>b=3
%28x%5E2++-+4x%2B2%5E2%29-2%5E2%2B+%28y%5E2+-+6y+%2B3%5E2%29-3%5E2%2B+9+=+0
%28x++-+2%29%5E2-4%2B+%28y+-+3%29%5E2-9%2B+9+=+0
%28x++-+2%29%5E2%2B+%28y+-+3%29%5E2-4+=+0
%28x++-+2%29%5E2%2B+%28y+-+3%29%5E2=+4
as you can see, h=2,k=3, and r=2
so, the center is at (2,3)
now we have two points, (5,5) and (2,3), and we can find the equation of a line passing through these two points

Solved by pluggable solver: Find the equation of line going through points
hahaWe are trying to find equation of form y=ax+b, where a is slope, and b is intercept, which passes through points (x1, y1) = (5, 5) and (x2, y2) = (2, 3).
Slope a is .
Intercept is found from equation a%2Ax%5B1%5D%2Bb+=+y%5B1%5D, or 0.666666666666667%2A5+%2Bb+=+1.66666666666667. From that,
intercept b is b=y%5B1%5D-a%2Ax%5B1%5D, or b=5-0.666666666666667%2A5+=+1.66666666666667.

y=(0.666666666666667)x + (1.66666666666667)

Your graph:




since your line is y=0.666666666666667x+%2B+1.66666666666667
or y=%282%2F3%29x+%2B+5%2F3, we can multiply both sides by 3
3y=3%282%2F3%29x+%2B+3%285%2F3%29
3y=2x+%2B+5, in standard form will be
2x-3y%2B5=0
so, your answer is:
d. 2x+-+3y+%2B+5+=+0





Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
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 
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) = %28matrix%281%2C3%2C-D%2F2%2C%22%2C%22%2C-E%2F2%29%29

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 %283-5%29%2F%282-5%29=%28-2%29%2F%28-3%29=2%2F3

And equation:  y-3 = 2%2F3(x-2)
              3y-9 = 2(x-2)
              3y-9 = 2x-4
          -2x+3y-5 = 0
           2x-3y+5 = 0

Choice d. 

Edwin