Chapter 12, Problem 4PS

a.

To determine

Determine the slope of the line joining $P$ and $O(0,0)$ ?

Answer to Problem 4PS

  $\boxed{\frac{4}{3}}$

Explanation of Solution

Given information:

Let $P(3,4)$ be a point on the circle $x^2+y^2=25$ see figure

  

What is the slope of the line joining $P$ and $O(0,0)$ ?

Calculation:

Consider the following diagram

  

To determine the slope of the line joining $P$ and $O(0,0)$ ,proceed as follows.

Use the standard form of the slope joining two points.

Standard form:

  $slope=\frac{y_2-y_1}{x_2-x_1}$

Hence the slope

  $\frac{y_2-y_1}{x_2-x_1}=\frac{3-0}{4-0}$

  $=\boxed{\frac{4}{3}}$

b.

To determine

Find an equation of the tangent line to the circle at $P$ .

Answer to Problem 4PS

  $\boxed{4y+3x=25}$

Explanation of Solution

Given information:

Let $P(3,4)$ be a point on the circle $x^2+y^2=25$ see figure

  

Find an equation of the tangent line to the circle at $P$ .

Calculation:

Consider the following diagram

  

Find the slope of the line perpendicular to the points joining $P$ and $O$ using the formula.

Product of the slopes of the two perpendicular lines is $1$ .

Slope of the line perpendicular to $OP$ is $-\frac{3}{4}$

Now use the standard equation of the line passing through a point $\left(x_1,y_1\right)$ and slope $m$ .

  $y-y_1=m\left(x-x_1\right)$

Hence the equation is

  $\begin{array}{l}y-4=-\frac{3}{4}\left(x-3\right)\\ 4y-16=-3x+9\end{array}$

Hence, $\boxed{4y+3x=25}$

c.

To determine

Find the slope $m_x$ of the line joining $P$ and $Q$ in terms of $x$ .

Answer to Problem 4PS

  $\boxed{m_x=\frac{\sqrt{25-x^2}-4}{x-3}}$

Explanation of Solution

Given information:

Let $P(3,4)$ be a point on the circle $x^2+y^2=25$ see figure

  

Let $Q\left(x,y\right)$ be another point on the circle in the first quadrant. Find the slope $m_x$ of the line joining $P$ and $Q$ in terms of $x$ .

Calculation:

Consider the following diagram

  

Consider the point $Q\left(x,y\right)$ on the first quadrant.

To determine the slope $m_x$ of the line joining $P$ and in terms of $x$ , proceed as follows.

Use the standard form of the slope joining two points.

Standard form:

  $slope=\frac{y_2-y_1}{x_2-x_1}$

Hence the slope is

  $m_x=\frac{y-4}{y-3}$

Now express $y$ in terms of $x$ using the givem equation.

  $\begin{array}{l}{x}^2+y^2=25\\ y^2=25-x^2\end{array}$

  $y=\pm \sqrt{25-x^2}$

The point os on the first quadrant, hence the values of $x$ and $y$ are positive.

Hence,

  $\boxed{m_x=\frac{\sqrt{25-x^2}-4}{x-3}}$

d.

To determine

Evaluate $\underset{x\to 3}{\lim }{m}_x$

Answer to Problem 4PS

  $=\boxed{\frac{-3}{4}}$

Explanation of Solution

Given information:

Let $P(3,4)$ be a point on the circle $x^2+y^2=25$ see figure

  

Evaluate $\underset{x\to 3}{\lim }{m}_x$ . How does this number relate to your answer in part (b)?

Calculation:

Consider the following diagram

  

Identfy the function and use rationalization method to evaluate.

  $\underset{x\to 3}{\lim }{m}_x=\underset{x\to 3}{\lim }\frac{\sqrt{25-x^2}-4}{x-3}$

Rationalize with $\sqrt{25-x^2}+4$

  $=\underset{x\to 3}{\lim }\frac{\sqrt{25-x^2}-4}{x-3}\times \frac{\sqrt{25-x^2}+4}{\sqrt{25-x^2}+4}$

Use mathematical identity $\left(a+b\right)\left(a-b\right)=\left(a^2-b^2\right)$

  $=\underset{x\to 3}{\lim }\frac{-\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(\sqrt{25-x^2}+4\right)}$

  $=\underset{x\to 3}{\lim }\frac{-\left(x+3\right)}{\left(\sqrt{25-x^2}+4\right)}$

Further simplify,

  $\begin{array}{l}=\frac{-\left(3+3\right)}{\left(\sqrt{25-9}+4\right)}\\ =\frac{-6}{8}\end{array}$

  $=\boxed{\frac{-3}{4}}$