Question

Find an equation of the line containing the centers of the two circles$$\begin{gathered} x^2+y^2-4x+6y+4=0and \\ {x}^2+y^2+6x+4y+9=0 \end{gathered}$$

Short answer

The equation of the line containing the centers of two circles is$y=-\frac{1}{5}x-\frac{13}{5}$

Step-by-step solution

Step 1. Given information

Here equations of two circles are given.

we need to find out the equation of the line containing the centers of two circles

Step 2. Finding the standard form of the equation of  a circle x2+y2-4x+6y+4 =0 

Here equation of the circle is $x^2+y^2-4x+6y+4=0$. To get standard form , first group the x containing term and do the same for y , then $(x^2-4x)+(y^2+6y)=-4$. Now complete square inside the parenthesis ,

$$\begin{gathered} (x^2-4x+4)+(y^2+6y+9)=-4+4+9 \\ {(x-2)}^2+({(y+3)}^2=9Thisisthes\tan dardformofthegivenequationofacircle. \end{gathered}$$

. Find the standard form of the equation x2+y2+6x+4y+9=0 

To convert equation in standard from, first group the terms according to x and y containing and then square inside the parenthesis then we get standard form equation of the given circle ,

$$\begin{gathered} x^2+y^2+6x+4y+9=0 \\ (x^2+6x)+(y^2+4y)=-9 \\ (x^2+6x+9)+(y^2+4y+4)=-9+9+4 \\ {(x+3)}^2+{(y+2)}^2=4,thisistheequationofas\tan dardformofthegivencircle \end{gathered}$$

Step 4.   Finding the center of both circles

From the standard form of each circle we get center, Center of the circle $x^2+y^2-4x+6y+4=0$is (2, -3) and $x^2+y^2+6x+4y+9=0$is (-3,-2)

.Finding the slope 

To find the equation containing line first we need to know about the slope. So slope m between containing points of first circle (2,-3) and the second circle(-3,-2) is

$$\begin{gathered} m=\frac{(y_2-y_1)}{(x_2-x_1)} \\ =\frac{-2-(-3)}{-3-2} \\ =-\frac{1}{5} \end{gathered}$$

 Step 4. Formulation equation of  the line containing the centers of two circles

By using point-slope form of a line we get the equation of a line$$\begin{gathered} (y-y_1)=m(x-x_1) \\ (y-(-3\left)\right)=\frac{-1}{5}(x-2)here(x_1,y_1)=(2,-3)andslopem=-\frac{1}{5}andsolve \\ (y+3)=-\frac{1}{5}x+\frac{2}{5} \\ y=-\frac{1}{5}x+\frac{2}{5}-3 \\ y=-\frac{1}{5}x-\frac{13}{5},thisisequationofalinecontainingpointsofacircle \end{gathered}$$