SOLUTION: Find an equation of the line (in either general form or slope intercept form) that contains the centers of the following two circles: x^2+y^2-4x+6y+4 = 0 and x^2+y^2+

Algebra ->  Equations -> SOLUTION: Find an equation of the line (in either general form or slope intercept form) that contains the centers of the following two circles: x^2+y^2-4x+6y+4 = 0 and x^2+y^2+      Log On


   

Question 14354: Find an equation of the line (in either general form or slope intercept form) that contains the centers of the following two circles:
x^2+y^2-4x+6y+4 = 0 and x^2+y^2+6x+4y+9 = 0
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
The equation of a circle with centre as (h,k) and radius r is given by the formula
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2.
So we have to write the given equation in this form to get the centre of the circle.
x%5E2%2By%5E2-4%2Ax%2B6%2Ay%2B4=0 OR x%5E2-4%2Ax%2By%5E2%2B6%2Ay%2B4=0
%28x-2%29%5E2-4%2B%28y%2B3%29%5E2-9%2B4=0...Note that we have added and subtracted 4 and 9 to make up the squares mentioned in brackets.
Hence %28x-2%29%5E2%2B%28y%2B3%29%5E2=9
Hence (2,-3)is the centre of this circle .Similarly the centre of the second circle is obtained from
x%5E2%2By%5E2%2B6%2Ax%2B4%2Ay%2B9=0 OR x%5E2%2B6%2Ax%2By%5E2%2B4%2Ay%2B9=0
%28x%2B3%29%5E2-9%2B%28y%2B2%29%5E2-4%2B9=0...Note that we have added and subtracted 9 and 4 to make up the squares mentioned in brackets.
Hence %28x%2B3%29%5E2%2B%28y%2B2%29%5E2=4
Hence the centre of the second circle is (-3,-2)
The equation of line joining 2 ponts (x1,y1) and (x2,y2)is given by the formula
Y-y1=%28y2-y1%29%2A%28X-x1%29%2F%28%28x2-x1%29%29
Substituting (2,-3) and (-3,-2) in the above formula we get
Y-(-3)=(-2-(-3))*(X-2)/(-3-(-2))
Y+3=(X-2)/(-5)
-5Y-15=X-2
X+5Y+13=0