Chapter 1.10, Problem 64E

(a)

To determine

To find: the equation of tangent to the circle given below at a given point.

Answer to Problem 64E

The equation of blue line is $\text{y+4}=\frac{3}{4}\left(x-3\right)$ .

Explanation of Solution

Given:

The equation is;

  ${\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{=25}$ at the $\left(3,-4\right)$ .

Concept used:

Some formula for straight line:

  $\text{m=}\frac{{\text{y}}_{\text{2}}{\text{-y}}_{\text{1}}}{{\text{x}}_{\text{2}}{\text{-x}}_{\text{1}}}$ .

If two Slopes are perpendicular to each other:

  ${\text{m}}_{\text{1}}{\text{.m}}_{\text{2}}\text{=-1}\text{.}$

Calculation:

First find the slope of the line segment with endpoints $\left(0,0\right)$ and $\left(3,-4\right)$

By using the formula:

  $\left(3,-4\right)$

  $m=\frac{y_2-y_1}{x_2-x_1}$

  $\Rightarrow \text{m=}\frac{\text{-4-0}}{\text{3-0}}$ .

  $\Rightarrow \text{m=-}\frac{\text{4}}{\text{3}}$ .

As any line tangent to a circle at point $\text{A}\left({\text{x}}_{\text{0}}{\text{,y}}_{\text{0}}\right)$ is perpendicular to the radius of the circle at this point.

So, the blue line is the picture is perpendicular to the line segment with the slope $\text{m=-}\frac{4}{3}$ .hence the slope of this line is:

  ${\text{m}}_{\text{1}}\text{=-}\frac{\text{1}}{\frac{\text{-4}}{\text{3}}}\text{=}\frac{\text{3}}{\text{4}}$ .

Hence the equation of blue line is $\text{y+4}=\frac{3}{4}\left(x-3\right)$ .

(b)

To determine

To find:the other point on circle where tangent line is parallel to another tangent line.

Answer to Problem 64E

The point is $\text{<msup> <mo>A </mo><mo>′</mo></msup>}\left(\text{-3,4}\right)$ .

Explanation of Solution

Given:

The equation is;

  ${\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}\text{=25}$ at the $\left(3,-4\right)$ .

Concept used:

The point slope form of linear equations.

  ${\text{y-y}}_{\text{0}}\text{=m}\left({\text{x-x}}_{\text{0}}\right)$ .

If two line is parallel to each other.

  ${\text{m}}_{\text{1}}{\text{=m}}_{\text{2}}$ .

Calculation:

Finally find the equation of the line that passes through the point $\text{A}\left({\text{x}}_{\text{0}}{\text{,y}}_{\text{0}}\right)$ = $\left(3,-4\right)$ with the slope ${\text{m}}_{\text{1}}\text{=}\frac{\text{3}}{\text{4}}$ .

User the point slope form of linear equations.

  ${\text{y-y}}_{\text{0}}\text{=m}\left({\text{x-x}}_{\text{0}}\right)$ .

  $\Rightarrow \text{y-}\left(-4\right)\text{=}\frac{3}{4}\left(\text{x-3}\right)$ .

  

  $\Rightarrow \text{3x-4y-9-16=0}$ .

  $\Rightarrow \text{3x-4y-25=0}$ .

The tangent will be parallel to the line found above at the point A that is symmetrical to $\text{A}\left(\text{-3,4}\right)$ about the origin.

That is the point $\text{<msup><mo> A </mo><mo>′</mo></msup>}\left(\text{-3,4}\right)$ .