Chapter 13.3, Problem 48E

To determine

To find:The equation for the farthermost points of the water.

Answer to Problem 48E

The equation for the farthermost points of the water is $x^2+y^2-14x-4y-56=0$

Explanation of Solution

Given:

The points is,

  $\begin{array}{l}{P}_1\left(7,2\right)\\ P_2\left(10,-8\right)\end{array}$

Calculation:

Calculate the points.

  $\begin{array}{c}{P}_1{P}_2=\sqrt{{\left(10-7\right)}^2+{\left(-8-2\right)}^2}\\ =\sqrt{3^2+{10}^2}\\ =\sqrt{109}\end{array}$

Now, the required equation,

  $\begin{array}{l}{\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2=r^2\\ {\left(x-7\right)}^2+{\left(y-2\right)}^2=109\\ x^2+y^2-14x-4y-56=0\end{array}$

Thus, the equation for the farthermost points of the water is $x^2+y^2-14x-4y-56=0$