Chapter 13, Problem 23RP

To determine

To find: The distance between the centre of the circles represented by the given equations.

Answer to Problem 23RP

The distance between the two centre is $9.2$ .

Explanation of Solution

Given:

The given equation are $\left(x-5\right)+{\left(y-2\right)}^2=29$ and $x^2+8x+y^2=31$ .

Calculation:

Consider the given equation are,

  $\begin{array}{l}\left(x-5\right)+{\left(y-2\right)}^2=29\\ x^2+8x+y^2=31\end{array}$

Consider the second equation is of the form,

  $\begin{array}{l}{x}^2+8x+y^2=31\\ {\left(x+4\right)}^2+y^2=47\end{array}$

The general form for the equation of the circle is,

  ${\left(x^2-h\right)}^2+{\left(y-k\right)}^2=r^2$

Then, the centre of the circle for the equation $x^2+8x+y^2=31$ is $\left(-4,0\right)$ and for the equation $\left(x-5\right)+{\left(y-2\right)}^2=29$ is $\left(5,2\right)$ .

Consider the distance between the two points is of the form,

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Then, distance between the two points of the circle is,

  $\begin{array}{c}d=\sqrt{{\left(5-\left(-4\right)\right)}^2+{\left(2-0\right)}^2}\\ =\sqrt{85}\\ =9.2\end{array}$