Find the distance between the centers of the circles x²+y²-2x+4y-11=0 and x²+y²+4x+2y-9=0.

The problem gives us the general formula of two circles so we're going to transform them into the center-radius formula to find the coordinates of the centers and then we'll find the distance between them.

The first circle is x²+y²-2x+4y-11=0

we are going to rearrange the terms of this equation and we get

(x² - 2x) + (y²+4y) = 11

we are going to complete it so the binomials become perfect trinomial squares:

(x²- 2x +1) + (y² + 4y + 4) = 11 + 1 + 4

(x-1)² + (y + 2)² = 16.

So this is a circle with center in (1, - 2) and radius 4.

We are going to do the same with the second circle:

x²+y²+4x+2y-9=0.

x² + 4x + y² + 2y = 9

(x²+ 4x + 4)² + (y² + 2y + 1)² = 9 +4 + 1

(x+2)² + (y + 1)² = 14

So this is a circle with center in (-2, 1) and radius √14

Now we need to find the distance between both centers which means finding the distance between the points (1, -2) and (-2, 1).

The formula for distance is given by:

[tex]d= \sqrt{(x_{2}-x_{1} )^{2} +(y_{2} - y_{1})^2 }[/tex]

Substituting our points into the formula we get:

[tex]d= \sqrt{(x_{2}-x_{1} )^{2} + (y_{2} - y_{1})^2 }\\d = \sqrt{(-2-1)^2+ (1-(-2)^2} \\d=\sqrt{(-3)^2+(3)^2}\\d=\sqrt{9+9} =\sqrt{18} = 4.242[/tex]

Therefore, the distance between the two centers is 4.242

erinna erinna · Answer 2 · 2019-11-22T11:42:55+01:00

Answer:

[tex]\sqrt{10}[/tex]

Step-by-step explanation:

The standard form of a circle is

[tex](x-h)^2+(y-k)^2=r^2[/tex] .... (1)

where, (h,k) is center of the circle and r is radius.

If a expression is [tex]x^2+bx[/tex], then we need to add [tex](\frac{b}{2})^2[/tex] to make it perfect square.

The equation of first circle is

[tex]x^2+y^2-2x+4y-11=0[/tex]

[tex](x^2-2x)+(y^2+4y)=11[/tex]

[tex](x^2-2x+1)-1+(y^2+4y+4)-4=11[/tex]

[tex](x-1)^2+(y-2)^2-5=11[/tex]

Add 5 on both sides.

[tex](x - 1)^2 + (y + 2)^2 = 16[/tex] .... (2)

On comparing (1) and (2) we get

[tex]h=1,k=-2[/tex]

The center of first circle is (1,-2).

The equation of second circle is

[tex]x^2+y^2+4x+2y-9=0[/tex]

[tex](x^2+4x)+(y^2+2y)=9[/tex]

[tex](x^2+4x+4)-4+(y^2+2y+1)-1=9[/tex]

[tex](x+2)^2+(y+1)^2-5=9[/tex]

Add 5 on both sides.

[tex](x + 2)^2 + (y + 1)^2 = 14[/tex] .... (3)

On comparing (1) and (3) we get

[tex]h=-2,k=-1[/tex]

The center of first circle is (-2,-1).

Distance formula:

[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

The distance between the centers of the circles

[tex]D=\sqrt{\left(-2-1\right)^2+\left(-1-\left(-2\right)\right)^2}=\sqrt{10}[/tex]

Therefore, the distance between the centers of the circles is [tex]\sqrt{10}[/tex].