Chapter 1.1, Problem 57E

To determine

To find: The equation of the line tangent to the circle at the point $\left(3,4\right)$ .

Answer to Problem 57E

The equation of the line tangent to the circle at the point $\left(3,4\right)$ is $y=-\frac{3}{4}x+\frac{25}{4}$ .

Explanation of Solution

Given information: The circle of the radius is $5$ and centered at $\left(0,0\right)$ .

Calculation:

The formula for the slope of the line that passes through the two points is given by,

  $m=\frac{y_2-y_1}{x_2-x_1}$

Substitute $0$ for $x_1$ , $0$ for $y_1$ , $3$ for $x_2$ , and $4$ for $y_2$ in the above formula.

  $\begin{array}{c}m=\frac{4-0}{3-0}\\ =\frac{4}{3}\end{array}$

The tangent line is tangent to the radius line, so it slope will be as follows.

  $\begin{array}{c}{m}^{\prime }=\frac{-1}{\frac{4}{3}}\\ =-\frac{3}{4}\end{array}$

The point slope equation of line is given by,

  $y-y_1=m\left(x-x_1\right)$

Substitute $-3$ for $x_1$ , $4$ for $y_1$ , and $-\frac{3}{4}$ for $m$ in the above formula.

  $\begin{array}{c}y-4=-\frac{3}{4}\left(x-3\right)\\ y=-\frac{3}{4}x+\frac{9}{4}+4\\ y=-\frac{3}{4}x+\frac{25}{4}\end{array}$

Therefore, the equation of the line tangent to the circle at the point $\left(3,4\right)$ is $y=-\frac{3}{4}x+\frac{25}{4}$ .