Chapter 12, Problem 2PS

a.

To determine

Slope of the given line.

Answer to Problem 2PS

Slope of the line joining $P$ and $O(0,0)$ is $\frac{4}{3}$ .

Explanation of Solution

Given Information: A circle whose equation is $x^2+y^2=25$ and a point on the circle is $P(3,4)$

Calculation:

Slope of the line joining the points $(x_1,y_1)and(x_2,y_2)$ is $m=\frac{y_2-y_1}{x_2-x_1}$ .

So, the slope of line joining $P(3,4)andO(0,0)$ is $m=\frac{4-0}{3-0}=\frac{4}{3}$ .

b.

To determine

An equation of the tangent line to the circle at $P$ .

Answer to Problem 2PS

The equation of tangent line to the circle at $P$ is $4y=-3x+25$ .

Explanation of Solution

Given Information: A circle whose equation is $x^2+y^2=25$ and a point on the circle is $P(3,4)$ .

Calculation:

The tangent line to the circle at $P$ is perpendicular to $OP$ .

From part (a), slope of normal $OP$ is $\frac{4}{3}$ .

So, slope of tangent $=\frac{-1}{4/3}=\frac{-3}{4}$ ( $\because$ Slope of tangent $\times$ Slope of normal $=-1$ )

So equation of tangent line is $y=\frac{-3}{4}x+c$ , where $c$ is the $y-\mathrm{int}ercept.$

Since tangent line passes through $(3,4)$ , so

  $\begin{array}{l}4=\frac{-3}{4}(3)+c\\ \Rightarrow c=4+\frac{9}{4}\\ \Rightarrow c=\frac{25}{4}\end{array}$

Thus, the equation of tangent line is $y=\frac{-3}{4}x+\frac{25}{4}$ , i.e., $4y=-3x+25$ .

c.

To determine

The slope of line joining $PandQ$ in terms of $x$ .

Answer to Problem 2PS

The slope of line joining $PandQ$ in terms of $x$ is: $m_x=\frac{4-\sqrt{25-x^2}}{3-x}$ .

Explanation of Solution

Given Information: A circle whose equation is $x^2+y^2=25$ and a points on the circle are $P(3,4)$ and $Q(x,y)$ .

Calculation:

Since the point $Q(x,y)$ lies on the circle, so it satisfies the equation of the circle, i.e.,

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

  $\Rightarrow y=\sqrt{25-x^2}$ ( $\because$ $Q$ lies in first quadrant, so $y$ can’t be negative)

So coordinates of $Q$ are $(x,\sqrt{25-x^2)}$ .

Thus, the slope of line joining $P(3,4)$ and $Q(x,\sqrt{25-x^2)}$ in terms of $x$ is:

  $m_x=\frac{4-\sqrt{25-x^2}}{3-x}$ .

d.

To determine

To evaluate: $\underset{x\to 3}{\lim }{m}_x$ and to relate this number to the answer in part (b).

Answer to Problem 2PS

  $\underset{x\to 3}{\lim }{m}_x=\frac{-3}{4}$ and it is the slope of the tangent line to the circle at $P$ .

Explanation of Solution

Given Information: A circle whose equation is $x^2+y^2=25$ and a point on the circle is $P(3,4)$ and result from part (b).

Calculation:

  $m_x=\frac{4-\sqrt{25-x^2}}{3-x}$

  $\Rightarrow \underset{x\to 3}{\lim }{m}_x=\underset{x\to 3}{\lim }\frac{4-\sqrt{25-x^2}}{3-x}$ ( $\frac{0}{0}$ form)

Let us rewrite the fraction by rationalizing the numerator.

  $\begin{array}{l}\frac{4-\sqrt{25-x^2}}{3-x}=(\frac{4-\sqrt{25-x^2}}{3-x})(\frac{4+\sqrt{25-x^2}}{4+\sqrt{25-x^2}})\\ =\frac{16-(25-x^2)}{(3-x)(4+\sqrt{25-x^2})}\\ =\frac{-(9-x^2)}{(3-x)(4+\sqrt{25-x^2})}\\ =\frac{-(3-x)(3+x)}{(3-x)(4+\sqrt{25-x^2})}\\ =\frac{-(3+x)}{4+\sqrt{25-x^2}}\\ \end{array}$

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

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

  $\begin{array}{l}=\frac{-(3+3)}{4+\sqrt{25-3^2}}\\ =\frac{-6}{4+4}\\ =\frac{-6}{8}\\ =\frac{-3}{4}\end{array}$

Thus, $\underset{x\to 3}{\lim }{m}_x=\frac{-3}{4}$

From part (b), the equation of tangent line to the circle at $P$ is $y=\frac{-3}{4}x+\frac{25}{4}$ .

So, $\underset{x\to 3}{\lim }{m}_x$ is the slope of the tangent line to the circle at $P$ .