Chapter 13.6, Problem 14PSB

(a)

To determine

To find: The equation for the tangent of the circle.

Answer to Problem 14PSB

The equation for the tangent of the circle is $y=\frac{-3}{4}x+\frac{25}{2}$ .

Explanation of Solution

Given:

The given point of tangent $\left(5,8\right)$ .

The given graph is shown in Figure 1

  

Figure 1

Calculation:

The slope of the line is calculated as,

  $\begin{array}{c}{m}_{PQ}=\frac{8-4}{6-3}\\ =\frac{4}{3}\end{array}$

Then, the slope of the tangent is,

  $-\frac{3}{4}$

The equation for the tangent is,

  $\begin{array}{l}y=-\frac{3}{4}x+b\\ 8=-\frac{3}{4}\left(6\right)+b\\ b=\frac{25}{1}\end{array}$

Thus, the tangent equation is,

  $y=\frac{-3}{4}x+\frac{25}{2}$

(b)

To determine

To find: The circumference of the circle.

Answer to Problem 14PSB

The circumference of the circle $10\pi$ .

Explanation of Solution

The radius of the circle by the distance formula is obtained as,

  $\begin{array}{c}r=\sqrt{{\left(6-3\right)}^2+{\left(8-4\right)}^2}\\ =5\end{array}$

Then, the circumference of the circle is,

  $\begin{array}{c}C=2\pi r\\ =2\left(5\right)\pi \\ =10\pi \end{array}$

(c)

To determine

To find: The distance of the point from A to Q.

Answer to Problem 14PSB

The value of AQ is $\sqrt{53}$ .

Explanation of Solution

From the figure the coordinates of point A is $\left(10,2\right)$ and that of Q is $\left(3,4\right)$ .

Then, by the distance formula the distance between both the points is obtained as,

  $\begin{array}{c}AQ=\sqrt{{\left(10-3\right)}^2+{\left(2-4\right)}^2}\\ =\sqrt{49+4}\\ =\sqrt{53}\end{array}$

(d)

To determine

To find: The distance of the point from A to the circle.

Answer to Problem 14PSB

The distance of the point from A to the circle $2.3$ .

Explanation of Solution

From the distance of the point from A to the circle is the difference between AQ and the radius of the circle as,

  $\begin{array}{c}AQ-r=\sqrt{53}-5\\ =2.3\end{array}$

(e)

To determine

To find: The area of the sector.

Answer to Problem 14PSB

The area of the sector is $13.1$ .

Explanation of Solution

Consider the formula to determine the area of the sector is,

  $A\left(\frac{60°}{360°}\right)$

Then,

  $\begin{array}{c}A\left(\frac{60°}{360°}\right)=\pi r^2\left(\frac{1}{6}\right)\\ =\frac{25\pi }{6}\\ =13.1\end{array}$