Chapter 13.7, Problem 16PSB

(a)

To determine

To find: The equation of the circle.

Answer to Problem 16PSB

The equation of the circle is ${\left(x-7\right)}^2+{\left(y-1\right)}^2=100$ .

Explanation of Solution

Given:

The given point is $\left(13,9\right)$ on the circle and the centre of the circle is at $\left(7,1\right)$ .

Calculation:

From the given points the radius of the circle is obtained as,

  $\begin{array}{c}r=\sqrt{{\left(13-7\right)}^2-{\left(9-1\right)}^2}\\ =\sqrt{36+64}\\ =10\end{array}$

Then, the equation of the circle is,

  ${\left(x-7\right)}^2+{\left(y-1\right)}^2=100$

(b)

To determine

To find: The area of the circle.

Answer to Problem 16PSB

The area of the circle is $100\pi$ .

Explanation of Solution

Consider the radius of the circle is 10.

Then, the area of the circle is,

  $\begin{array}{c}A=\pi r^2\\ =\pi {\left(10\right)}^2\\ =100\pi \end{array}$

(c)

To determine

To find: The circumference of the circle.

Answer to Problem 16PSB

The circumference of the circle is $20\pi$ .

Explanation of Solution

Consider the radius of the circle is 10.

Then, the circumference of the circle is,

  $\begin{array}{c}C=2\pi r\\ =2\pi \left(10\right)\\ =20\pi \end{array}$

(d)

To determine

To find: The coordinate of the point on the circle directly opposite $\left(13,9\right)$ .

Answer to Problem 16PSB

The coordinate of the points is $\left(1,-7\right)$ .

Explanation of Solution

Consider that $\left(7,1\right)$ is the midpoint of the points $\left(13,9\right)$ and $\left(x,y\right)$ .

Then,

  $\begin{array}{l}\left(\frac{13+x}{2},\frac{9+y}{2}\right)=\left(7,1\right)\\ x=1\\ y=-7\end{array}$

Thus, the coordinate of the points is $\left(1,-7\right)$ .

(e)

To determine

To find: The point slope form an equation of the line tangent to the circle at $\left(13,9\right)$ .

Answer to Problem 16PSB

The required equation is $y-9=\frac{3}{4}\left(x-13\right)$ .

Explanation of Solution

Consider the slope of the tangent is,

  $\begin{array}{c}m=\frac{9-1}{13-7}\\ =\frac{8}{6}\\ =\frac{4}{3}\end{array}$

The point slope form of the tangent to the point $\left(13,9\right)$ is obtained as,

  $y-9=\frac{3}{4}\left(x-13\right)$

(f)

To determine

To find: The distance between the centre of the circle and the $\left(19,6\right)$ .

Answer to Problem 16PSB

The required distance is $13$ .

Explanation of Solution

The distance between $\left(19,6\right)$ and the centre of the circle is of the form,

  $\begin{array}{c}\sqrt{{\left(19-7\right)}^2+{\left(6-1\right)}^2}=\sqrt{144+25}\\ =13\end{array}$

(g)

To determine

To find: The distance between the point $\left(19,6\right)$ and the circle.

Answer to Problem 16PSB

The required distance is $3$ .

Explanation of Solution

The distance between $\left(19,6\right)$ and the circle is obtained as the difference of distance from the centre and the radius.

  $\begin{array}{c}d=13-10\\ =3\end{array}$