Chapter 9, Problem 7CR

Solution

To find: $PZ$

  $PZ=10$

Given:

The figure:

Radius of the circle with center P is 6.

  $XZ=8$

Explanation:

In the given figure draw $\overline{PX}$ and $\overline{PZ}$ :

It can be observed that $\overline{XZ}$ is a tangent to circle with point of tangency X .

So, by theorem 9-1, it can be said that the radius $\overline{PX}$ is perpendicular to $\overline{XZ}$ .

That is,

  $m\angle PXZ=90$

Let $PZ=x$

Consider the right triangle $△PXZ$ .

Now, $\overline{PZ}$ is the hypotenuse and has length x and the lengths of the legs are $PX=6$ and $XZ=8$ .

So, by theorem 8-2,

  $\begin{array}{c}P{Z}^2=P{X}^2+X{Z}^2\\ x^2=6^2+8^2\\ x^2=36+64\\ x^2=100\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{100}\\ x=10\end{array}$

So, $PZ=10$