91Ó°ÊÓ

Here equations of two circles are given.

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

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 ,

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

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 ,

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

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)

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

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

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