Chapter 12, Problem 13CR

a.

To determine

To calculate:Thevalue ofAC with circle of center O.

Answer to Problem 13CR

The value ofACis$25$ .

Explanation of Solution

Given information:

A circle with center O and tangent $\overline{CD}$ .

  $\begin{array}{l}CD=15,\\ BC=9.\end{array}$

Calculation:

  

We use the below theorem:

Tangent-Secant Power Theorem: If a tangent and secant are drawn from an external point to a circle, then the square of the length of the tangent is equal to the product of the length of the secant’s external part and the length of the entire secant.

By applyingTangent-Secant Power Theorem, we get

  $\begin{array}{l}{\left(CD\right)}^2=\left(BC\right)\left(AC\right)\\ {\left(15\right)}^2=\left(9\right)\left(AC\right)\\ 225=\left(9\right)\left(AC\right)\\ AC=\frac{225}{9}\\ AC=25\end{array}$

b.

To determine

To find: The value of diameter AB with circle of center O.

Answer to Problem 13CR

The value of diameterAB is $16$ .

Explanation of Solution

Given information:

A circle with center O and tangent $\overline{CD}$ .

  $\begin{array}{l}CD=15,\\ BC=9.\end{array}$

Calculation:

  

We use the below theorem:

Tangent-Secant Power Theorem: If a tangent and secant are drawn from an external point to a circle, then the square of the length of the tangent is equal to the product of the length of the secant’s external part and the length of the entire secant.

By applyingTangent-Secant Power Theorem, we get

  $\begin{array}{l}{\left(CD\right)}^2=\left(BC\right)\left(AC\right)\\ {\left(15\right)}^2=\left(9\right)\left(AC\right)\\ 225=\left(9\right)\left(AC\right)\\ AC=\frac{225}{9}\\ AC=25\end{array}$

  $\begin{array}{l}AB=AC-BC\\ AB=25-9=16\end{array}$