Chapter 13.7, Problem 14CE

a.

To determine

To find:the radius of the circle

Answer to Problem 14CE

The radius of the circle is 5

Explanation of Solution

Given information:

Line $l$ is length to $\odot O$ at point $P\left(3,4\right)$

  

Formula used:

  $r=\sqrt{{\left(x-h\right)}^2+{\left(y-k\right)}^2}\text{ where, }h,k\text{ be the point of the center of the circle}$

Calculation:

Let $r$ be the radius of the circle

Let $\left(0,0\right)$ be the point of the center of the circle

The distance from the center to the point of the tangent

  $\begin{array}{l}r=\sqrt{{\left(3-0\right)}^2+{\left(4-0\right)}^2}\\ r=\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}\\ r=\sqrt{9+16}\\ r=\sqrt{25}\\ r=5\end{array}$

Therefore, $r=5$

Hence, the radius of the circle is 5.

b.

To determine

To find:an equation of the circle

Answer to Problem 14CE

The equation of the circle is $x^2+y^2=25$

Explanation of Solution

Given information:

Line $l$ is length to $\odot O$ at point $P\left(3,4\right)$

  

Formula used:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\text{ where, }h,k\text{ be the point of the center of the circle}$

Calculation:

Let $r$ be the radius of the circle

Let $\left(0,0\right)$ be the point of the center of the circle

The radius of the circle is 5

  $\begin{array}{l}{\left(x-0\right)}^2+{\left(y-0\right)}^2=5^2\\ x^2+y^2=25\end{array}$

Hence, the equation of the circle is $x^2+y^2=25$ .

c.

To determine

To find: the slope of line $l$

Answer to Problem 14CE

The slope of the line $l$ is $-\frac{3}{4}$

Explanation of Solution

Given information:

Line $l$ is length to $\odot O$ at point $P\left(3,4\right)$

  

Formula used:

  $\text{the slope of the line}=\frac{y-y_1}{x-x_1}$

Calculation:

The slope of the line

  $\begin{array}{l}OP=\frac{4-0}{3-0}\\ OP=\frac{4}{3}=m_2\end{array}$

Let $m_1$ be the slope of the line $l$ .

The line $l$ and $OP$ are perpendicular

So,

  $\begin{array}{l}{m}_1\cdot m_2=-1\\ m_1\cdot \frac{4}{3}=-1\\ m_1=-\frac{3}{4}\end{array}$

Hence, the slope of the line $l$ is $-\frac{3}{4}$ .

d.

To determine

To find:the equation of the line $l$

Answer to Problem 14CE

The equation of the line $l$ is $3x+4y=25$

Explanation of Solution

Given information:

Line $l$ is length to $\odot O$ at point $P\left(3,4\right)$

  

Formula used:

  $y=mx+b$

Calculation:

Let $m$ be the slope of line $l$

Therefore,

  $m=-\frac{3}{4}$

The equation of the $l$

  $y=-\frac{3}{4}x+b$

The point $\left(3,4\right)$ lies on it

  $\begin{array}{l}4=-\frac{3}{4}\cdot 3+b\\ 4=-\frac{9}{4}+b\\ b=\frac{9}{4}+4\\ b=\frac{25}{4}\end{array}$

Therefore, the equation of the line $l$ is

  $\begin{array}{l}y=-\frac{3}{4}x+\frac{25}{4}\\ 4y=-3x+25\\ 3x+4y=25\end{array}$

Hence, the equation of the line $l$ is $3x+4y=25$ .