Question

Find an equation of the line containing the centers of the two circles whose equations are given below.

Answer 1

Answer:

3y+x = -5

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, we will compare both equations with the general form of the equation above as shown;

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

Hence the equation of the line containing the centers of the two circles is 3y+x = -5