Chapter 8.3, Problem 54E

Solution

To write: the equation of line tangent to the given circle at the given point.

  $y=\frac{2}{3}x-\frac{13}{3}$

Given information:

The given circle and the point is:

  $x^2+y^2=13;\left(2,-3\right)$

Formula Used:

Standard equation of a circle with Center at the Origin

The standard form of the equation of a circle with center at $\left(0,0\right)$ and radius r is as follows:

  $x^2+y^2=r^2$

Slope of the line segment

The slope of the line segment joining the points $A\left(x_1,y_1\right)$ and $B\left(x_2,y_2\right)$ is:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Perpendicular lines and their slopes

If two lines are perpendicular, then their slopes are negative reciprocals of each other.

That is, if $m_1$ and $m_2$ are the slopes of two perpendicular lines, then

  $m_2=-\frac{1}{m_1}$

Equation of the line

The equation of a line passing through $\left(x_0,y_0\right)$ and having the slope m is:

  $y-y_0=m\left(x-x_0\right)$

Explanation:

Now, the given equation of the circle $x^2+y^2=13$ is the standard equation of circle with center at origin. Also, the given point $\left(2,-3\right)$ lies on the circle. So, the radius of the circle is the line joining $\left(0,0\right)$ and the point $\left(2,-3\right)$ .

So, first step is to find the slope that goes through $\left(0,0\right)$ and the point $\left(2,-3\right)$ .

  $\begin{array}{c}{m}_1=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{0+3}{0-2}\\ =-\frac{3}{2}\end{array}$

A line tangent to the given circle is perpendicular to the radius of the circle at the point of tangency.

So, the slope of the tangent line is negative reciprocal of the above slope.

  $\begin{array}{c}{m}_2=-\frac{1}{m_1}\\ =\frac{2}{3}\end{array}$

Then, the equation of the tangent line that passes through the point $\left(2,-3\right)$ and has the slope $\frac{2}{3}$ is:

  $\begin{array}{c}y-y_0=m\left(x-x_0\right)\\ y-\left(-3\right)=\frac{2}{3}\left(x-2\right)\\ y+3=\frac{2}{3}x-\frac{4}{3}\\ y=\frac{2}{3}x-\frac{4}{3}-3\\ y=\frac{2}{3}x-\frac{13}{3}\end{array}$