Chapter 8.3, Problem 55E

Solution

Write an equation of the line tangent to the given circle at the given point.

  $y=\frac{5}{3}x+\frac{34}{3}$

Given information:

Circle at $x^2+y^2=34$ .and tangent at $\left(-5,3\right)$

Formula used:

• Slope of a line between two points $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$

: $m=\frac{y_1-y_2}{x_1-x_2}$

• Slope of the tangent of a line with slope m is: $\left(-\frac{1}{m}\right)$
• Point slope formula of a line with slope m, and $\left(x_1,y_1\right)$ is

  $y-y_1=m\left(x-x_1\right)$

Calculation:

In order to write an equation of line tangent to the circle $x^2+y^2=34$ at the point $\left(-5,3\right)$

First find the slope of the line passing through center of circle and the given point $\left(-5,3\right)$ using the formula,

  $m=\frac{y_1-y_2}{x_1-x_2}$

As,

  $\begin{array}{c}m=\frac{y_1-y_2}{x_1-x_2}\\ =\frac{3-0}{-5-0}\\ =-\frac{3}{5}\end{array}$

Now, the slope of the line tangent to the circle at the point $\left(-5,3\right)$ will be perpendicular to the line passing through center of the circle and point $\left(-5,3\right)$ , so it’s slope will be

  $\begin{array}{c}\frac{-1}{m}=-\left(\frac{1}{-\frac{3}{5}}\right)\\ =-\left(-\frac{5}{3}\right)\\ =\frac{5}{3}\end{array}$

Now, using the point slope formula of a line with slope passing through point $\left(x_1,y_1\right)$

  $y-y_1=m\left(x-x_1\right)$

As,

  $\begin{array}{c}y-3=\frac{5}{3}\left(x-\left(-5\right)\right)\\ y-3=\frac{5}{3}\left(x+5\right)\\ y=\frac{5}{3}x+\frac{5\times 5}{3}+3\\ y=\frac{5}{3}x+\frac{25}{3}+3\\ y=\frac{5}{3}x+\frac{25+9}{3}\\ y=\frac{5}{3}x+\frac{34}{3}\end{array}$

Thus the required equation is

  $y=\frac{5}{3}x+\frac{34}{3}$