Chapter 1.4, Problem 56AYU

To determine

To find: an equation of the line.

Answer to Problem 56AYU

Thus, the equation of line is $y+3=-\frac{1}{5}\left(x-2\right)$ .

Explanation of Solution

Given:

  $x^2+y^2-4x+6y+4=0$

  $x^2+y^2+6x+4y+9=0$

The standard form of an equation of a circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ , where $r$ is the radius and $\left(h,k\right)$ is the center of the circle.

In order to convert the first equation to standard form, complete the square in both $x$ and $y$ .

Subtract $4$ from each side of the equation to isolate constant to one side.

  $\begin{array}{l}{x}^2+y^2-4x+6y+4-4=0-4\\ x^2+y^2-4x+6y=-4\end{array}$

Next, group the expressions involving $x$ and $y$ terms.

  $\left(x^2-4x\right)+\left(y^2+6y\right)=-4$

In order to complete the squares, take half the coefficients of $x$ and $y$ . Square the numbers and add them on both the sides of the equation.

  $\begin{array}{l}\left(x^2-4x+4\right)+\left(y^2+6y+9\right)=-4+4+9\\ \stackrel{\uparrow }{\overbrace{{\left(\frac{4}{2}\right)}^2=4}} \stackrel{\uparrow }{\overbrace{{\left(\frac{6}{2}\right)}^2=9}}\end{array}$

Factor the equation.

  ${\left(x-2\right)}^2+{\left(y+3\right)}^2=3^2$

The equation obtained can be rewritten ${\left(x-2\right)}^2+{\left[y-\left(-3\right)\right]}^2=3^2$ .

On comparing the equation with the standard form, the values obtained are $h$ as $2$ , $k$ as $-3$ and $r$ as $3$ .

Thus the center $\left(h,k\right)$ of the circle is $\left(2,-3\right)$ and the radius is $3$ .

Similarly, convert the second equation to standard form, complete the square in both $x$ and $y$ .

Subtract $9$ from each side of the equation to isolate constant to one side.

  $\begin{array}{l}{x}^2+y^2+6x+4y+9-9=0-9\\ x^2+y^2+6x+4y=-9\end{array}$

Next, group the expressions involving $x$ and $y$ terms.

  $\left(x^2+6x\right)+\left(y^2+4y\right)=-9$

In order to complete the squares, take half the coefficients of $x$ and $y$ . Square the numbers and add them on both the sides of the equation.

  $\begin{array}{l}\left(x^2+6x+9\right)+\left(y^2+4y+4\right)=-9+9+4\\ \stackrel{\uparrow }{\overbrace{{\left(\frac{6}{2}\right)}^2=3}} \stackrel{\uparrow }{\overbrace{{\left(\frac{4}{2}\right)}^2=4}}\end{array}$

Factor the equation.

  ${\left(x+3\right)}^2+{\left(y+2\right)}^2=2^2$

On comparing the equation with the standard form, the values obtained are $h$ as $-3$ , $k$ as $-2$ and $r$ as $2$ .

Thus the center $\left(h,k\right)$ of the circle is $\left(-3,-2\right)$ and the radius is $2$ .

In order to find the equation of the line passing through the centers of the two circles, use the point-slope form, $y-y_1=m\left(x-x_1\right)$

Find the slope

Substitute $2$ for $x_1$ , $-3$ for $y_1$ , $-3$ for $x_2$ , and $-2$ for $y_2$ in the formula, $m=\frac{y_2-y_1}{x_2-x_1}$

  $m=\frac{-2-\left(-3\right)}{-3-2}$

Simplify,

  $\frac{-2+3}{-3-2}=-\frac{1}{5}$

Substitute the values of $x_1,y_1$ , and $m$ in the point-slope form.

  $\begin{array}{l}y-\left(-3\right)=-\frac{1}{5}\left(x-2\right)\\ y+3=-\frac{1}{5}\left(x-2\right)\end{array}$

Thus, the equation of line is $y+3=-\frac{1}{5}\left(x-2\right)$ .

Conclusion:

Thus, the equation of line is $y+3=-\frac{1}{5}\left(x-2\right)$ .