Q.56 - Precalculus Enhanced with Graphing Utilities

Question

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

Step-by-Step Solution

Answer

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

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 x 2 + y 2 - 4 x + 6 y + 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 ,

$$\begin{gathered} (x^2-4x+4)+(y^2+6y+9)=-4+4+9 \\ {(x-2)}^2+({(y+3)}^2=9 This is the s\tan dard form of the given equation of a circle. \end{gathered}$$

Step 3 . Find the standard form of the equation x 2 + y 2 + 6 x + 4 y + 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 , 

$$\begin{gathered} x^2+y^2+6x+4y+9=0 \\ (x^2+6x)+(y^2+4y)=-9 \\ (x^2+6x+9)+(y^2+4y+4)=-9+9+4 \\ {(x+3)}^2+{(y+2)}^2=4, this is the equation of a s\tan dard form of the given circle \end{gathered}$$

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)  

Step 5 .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 $$\begin{gathered} (y-y_1)=m(x-x_1) \\ (y-(-3\left)\right)=\frac{-1}{5}(x-2) here (x_1,y_1)=(2,-3) and slope m=-\frac{1}{5} and solve \\ (y+3)=-\frac{1}{5}x+\frac{2}{5} \\ y=-\frac{1}{5}x+\frac{2}{5}-3 \\ y=-\frac{1}{5}x-\frac{13}{5} , this is equation of a line containing points of a circle \end{gathered}$$