Chapter 9.2, Problem 47WE

To determine

To find: The path of boat.

Answer to Problem 47WE

The boat’s path is circular that has a radius of 2 and center at $\left(0,4\right)$ .

Explanation of Solution

Given information: One buoy is at $\left(0,0\right)$ and other one is at $\left(3,0\right)$ .

Formula used: $D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation: The path followed by the boat is going to be circular. We have to set up an equation using the given distances from the point $\left(x,y\right)$ and the two buoys, to find the center and the radius of the circular path.

  $\begin{array}{l}\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}=2\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ \sqrt{x^2+y^2}=2\sqrt{{\left(x-3\right)}^2+y^2}\\ x^2+y^2=4x^2-24x+36+4y^2\\ 4x^2-x^2-24x+4y^2-y^2=-36\\ 3x^2-24x+3y^2=-36\\ 3\left(x^2-8x+16\right)+3y^2=-36+48\\ 3{\left(x-4\right)}^2+3y^2=12\\ {\left(x-4\right)}^2+y^2=4\end{array}$

Since the standard equation of the circle ${\left(x-h\right)}^2{\left(y-k\right)}^2=r^2$ has center $\left(h,k\right)$ and radius $\left(r\right)$ .

Compare ${\left(x-4\right)}^2+y^2=4$ to ${\left(x-h\right)}^2{\left(y-k\right)}^2=r^2$

Hence,

  $\begin{array}{l}\left(h,k\right)=\left(4,0\right)\\ r=2\end{array}$

Therefore, the path of the boat is a circle that has a radius of 2 and center at $\left(0,4\right)$ .