Chapter 10.8, Problem 20PSC

a)

To determine

To find : Diameter of the circle.

Answer to Problem 20PSC

  $26.25$

Explanation of Solution

Given Information :

Tangent AT measures, $12$ , AB = $8$ and AT $\perp$ AB.

  

Formula Used:

  $\frac{\text{diameter}}{\text{2}}$ = radius

Calculation :

Let, Radius = $r$

OT = $r$ = OM + MT

  $\begin{array}{l}\Rightarrow r=\text{OM}+8\\ \Rightarrow \text{OM}=r-8\\ \therefore \text{OB}=r\\ \because {\text{OM}}^2+{\text{MB}}^2={\text{OB}}^2\\ $ \Rightarrow {(r-8)}^2+{12}^2=r^2$\Rightarrow r^2-16r+64+144=r^2\\ \Rightarrow 210=16r\\ \text{So }r=\frac{105}{8}=0\\ \text{Diameter}=2r=\frac{105}{4}=26.25\end{array}$

b)

To determine

To find : How far is the circle from point A.

Answer to Problem 20PSC

  $17.784$

Explanation of Solution

Given Information :

The given information is MT = BT = $8$ , MB = TA = $12$

Formula Used:

Pythagoras Theorem : (hypotenuse)2 = (perpendicular)2 + (base)2

Calculation :

  $\begin{array}{l}{\text{OA}}^2={\text{OT}}^2+{\text{TA}}^2\\ =r^2+{(12)}^2\\ ={\left(\frac{210}{16}\right)}^2+144\\ \Rightarrow \text{OA}=\sqrt{{\left(\frac{210}{16}\right)}^2+144}=17.784\end{array}$