Chapter 9 Quadratic Relations and Conic Sections - 9.3 Graph and Write Equations of Circles - 9.3 Exercises - Problem Solving - Page 631: 65c

$6$

Let it enter the layer at $(x_2,-4)$ and leave at $(x_1,-4)$. As we can see $x_1=-x_2$ Then $x_1^2+(-4)^2=5^2\\x_1^2=9\\x_1=\pm3$ As we can see $x_1$ is positive, thus $x_1=3$. Thus $x_2=-3$. Then the distance is: $|x_1-x_2|=|3-(-3)|=6$