Chapter 9.7, Problem 22WE

(a)

To determine

To Find: the measure of chord $AB$ in the circle shown.

Answer to Problem 22WE

  $6$

Explanation of Solution

Given:

Circle is shown below.

  

  $\overrightarrow{BA}$ is perpendicular to $\overrightarrow{PT}\text{ at }P$ , $TA=6\text{ and }PA=3$ .

Calculation:

Since, $\Delta TPA$ is a right triangle therefore.

  $\begin{array}{l}T{A}^2=T{P}^2+P{A}^2\\ 6^2=T{P}^2+3^2\\ T{P}^2=36-9\\ T{P}^2=27\\ TP=3\sqrt{3}\end{array}$

Since, TP and PB are tangent and secant therefore.

  $\begin{array}{l}T{P}^2=PB\times PA\\ {\left(3\sqrt{3}\right)}^2=\left(PA+AB\right)\times PA\\ 27=\left(3+AB\right)\times 3\\ 3+AB=9\\ AB=6\end{array}$

Therefore, the answer is $6$ .

(b)

To determine

To Find: the distance from O to AB in the circle shown.

Answer to Problem 22WE

  $3\sqrt{3}$

Explanation of Solution

Given:

Circle is shown below.

  

  $\overrightarrow{BA}$ is perpendicular to $\overrightarrow{PT}\text{ at }P$ , $TA=6\text{ and }PA=3$ .

Calculation:

Since, $\Delta TPA$ is a right triangle therefore.

  $\begin{array}{l}T{A}^2=T{P}^2+P{A}^2\\ 6^2=T{P}^2+3^2\\ T{P}^2=36-9\\ T{P}^2=27\\ TP=3\sqrt{3}\end{array}$

Since, TP and PB are tangent and secant therefore.

  $\begin{array}{l}T{P}^2=PB\times PA\\ {\left(3\sqrt{3}\right)}^2=\left(PA+AB\right)\times PA\\ 27=\left(3+AB\right)\times 3\\ 3+AB=9\\ AB=6\end{array}$

Perpendicular from centre O to TP at T and AB is drawn below.

  

Here, OTPD is a rectangle and

  $\begin{array}{l}TP=OD\\ OD=3\sqrt{3}\end{array}$

Therefore, the answer is $3\sqrt{3}$ .

(c)

To determine

To Find: the radius of circle shown in figure.

Answer to Problem 22WE

  $6$

Explanation of Solution

Given:

Circle is shown below.

  

  $\overrightarrow{BA}$ is perpendicular to $\overrightarrow{PT}\text{ at }P$ , $TA=6\text{ and }PA=3$ .

Calculation:

Since, $\Delta TPA$ is a right triangle therefore.

  $\begin{array}{l}T{A}^2=T{P}^2+P{A}^2\\ 6^2=T{P}^2+3^2\\ T{P}^2=36-9\\ T{P}^2=27\\ TP=3\sqrt{3}\end{array}$

Since, TP and PB are tangent and secant therefore.

  $\begin{array}{l}T{P}^2=PB\times PA\\ {\left(3\sqrt{3}\right)}^2=\left(PA+AB\right)\times PA\\ 27=\left(3+AB\right)\times 3\\ 3+AB=9\\ AB=6\end{array}$

Perpendicular from centre O to TP at T and AB is drawn below.

  

Here, OTPD is a rectangle and

  $\begin{array}{l}TP=OD\\ OD=3\sqrt{3}\end{array}$

Radius is calculated as.

  $\begin{array}{l}OT=DP\\ r=OT\\ =AD+AP\\ =\frac{AB}{2}+AP\therefore \left(OD\text{ Bisects }AB\right)\\ =\frac{6}{2}+3\\ =6\end{array}$

Therefore, the answer is $6$ .