Chapter 9.7, Problem 27WE

To determine

To Find: the radius of circle shown in figure.

Answer to Problem 27WE

  $2\sqrt{10}$

Explanation of Solution

Given:

Figure is shown below.

  

Calculation:

Extend N to Q, ON to R and S as shown below.

  

Using property of secant and tangent.

  $\begin{array}{l}P{T}^2=PJ\times PQ\\ {12}^2=8\times \left(8+4+NQ\right)\\ 144=8\times \left(12+NQ\right)\\ NQ=18-12\\ NQ=6\end{array}$

A theorem state that “When two chords intersect inside a circle, the product of the segments of one chord equals the product of the segments of the other chord”.

  $\begin{array}{l}SN\times NR=NJ\times NQ\\ \left(r+4\right)\times \left(r-4\right)=4\times 6\\ r^2-4^2=24\\ r^2=24+16\\ r^2=40\\ r=\sqrt{40}\\ r=2\sqrt{10}\end{array}$

Therefore, the answer is $2\sqrt{10}$ .