Answer:
<h2><em>
3y+x = -5</em></h2>
Step-by-step explanation:
The general equation of a circle is expressed as x²+y²+2gx+2fy+c = 0 with centre at C (-g, -f).
Given the equation of the circles x²+y²−2x+4y+1 =0 and x²+y²+4x+2y+4 =0, to get the centre of both circles,<em> we will compare both equations with the general form of the equation above as shown;</em>
For the circle with equation x²+y²−2x+4y+1 =0:
2gx = -2x
2g = -2
Divide both sides by 2:
2g/2 = -2/2
g = -1
Also, 2fy = 4y
2f = 4
f = 2
The centre of the circle is (-(-1), -2) = (1, -2)
For the circle with equation x²+y²+4x+2y+4 =0:
2gx = 4x
2g = 4
Divide both sides by 2:
2g/2 = 4/2
g = 2
Also, 2fy = 2y
2f = 2
f = 1
The centre of the circle is (-2, -1)
Next is to find the equation of a line containing the two centres (1, -2) and (-2.-1).
The standard equation of a line is expressed as y = mx+c where;
m is the slope
c is the intercept
Slope m = Δy/Δx = y₂-y₁/x₂-x₁
from both centres, x₁= 1, y₁= -2, x₂ = -2 and y₂ = -1
m = -1-(-2)/-2-1
m = -1+2/-3
m = -1/3
The slope of the line is -1/3
To get the intercept c, we will substitute any of the points and the slope into the equation of the line above.
Substituting the point (-2, -1) and slope of -1/3 into the equation y = mx+c
-1 = -1/3(-2)+c
-1 = 2/3+c
c = -1-2/3
c = -5/3
Finally, we will substitute m = -1/3 and c = 05/3 into the equation y = mx+c.
y = -1/3 x + (-5/3)
y = -x/3-5/3
Multiply through by 3
3y = -x-5
3y+x = -5
<em>Hence the equation of the line containing the centers of the two circles is 3y+x = -5</em>
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