Chapter 13, Problem 18RP

(a)

To determine

To find: The equation of the circle P.

Answer to Problem 18RP

The equation of the circle is ${\left(x-3\right)}^2+{\left(y-2\right)}^2=40$ .

Explanation of Solution

Given:

The given diagram is shown in Figure 1

  

Figure 1

Calculation:

The general form for the equation of the circle is,

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

Then, the radius of the circle is obtained as,

  $\begin{array}{c}r=\sqrt{{\left(9-3\right)}^2+{\left(4-2\right)}^2}\\ =\sqrt{40}\end{array}$

Then, the equation of the circle is,

  ${\left(x-3\right)}^2+{\left(y-2\right)}^2=40$

(b)

To determine

To find: The area of the circle.

Answer to Problem 18RP

The area of the circle is $40\pi$ .

Explanation of Solution

Calculation:

Consider the radius of the circle is 40.

Then, the area of the circle is,

  $\begin{array}{c}A=\pi r^2\\ ={\left(\sqrt{40}\right)}^2\pi \\ =40\pi \end{array}$

(c)

To determine

To find: The coordinates of the V.

Answer to Problem 18RP

The coordinate of the $V=\left(-3,0\right)$ .

Explanation of Solution

The point P is the midpoint of VR when $V=\left(-3,0\right)$

(d)

To determine

To find: The point slope form of the equation of the tangent RT.

Answer to Problem 18RP

The equation of RT is $y-4=-3\left(x-9\right)$ .

Explanation of Solution

The formula to determine the slope of the line is of the form,

  $m=\frac{y_2-y_1}{x_2-x_1}$

Then, slope of the line RT is,

  $\begin{array}{c}m=\frac{4-\left(-14\right)}{9-15}\\ =\frac{18}{-6}\\ =-3\end{array}$

Then, the equation of the line is for the point $\left(9,4\right)$ is,

  $y-4=-3\left(x-9\right)$

(e)

To determine

To find: The value of PT.

Answer to Problem 18RP

The value of the PT is $20$ .

Explanation of Solution

The distance between the two points in the coordinate plane is of the form,

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Then, the length of the PT is,

  $\begin{array}{c}PT=\sqrt{{\left(15-3\right)}^2+{\left(-14-2\right)}^2}\\ =\sqrt{144+256}\\ =20\end{array}$

(f)

To determine

To find: The distance from T to the circle to nearest tenth.

Answer to Problem 18RP

The

Explanation of Solution

The distance between the two points in the coordinate plane is of the form,

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The distance from T to the circle is,

  $PT-r$

Then,

  $20-\sqrt{40}=13.7$

(g)

To determine

To find: The equation of the circle PRT.

Answer to Problem 18RP

The area of the triangle is $60$ .

Explanation of Solution

The area of the triangle $\Delta PRT$ .

Consider the area of the triangle is of the form,

  $A=\frac{1}{2}\left(PR\right)\left(RT\right)$

Then,

  $\begin{array}{c}A=\frac{1}{2}\left(\sqrt{40}\right)\left(\sqrt{360}\right)\\ =60\end{array}$