Chapter 9.1, Problem 108E

To determine

To find: The equation of the tangent line to the circle $x^2+y^2=169$ at the point $\left(-5,12\right)$ .

Answer to Problem 108E

The equation of the tangent line to the circle $x^2+y^2=169$ at the point $\left(-5,12\right)$ is $5x-12y+169=0$ .

Explanation of Solution

Given information:

The equation of the circle is $x^2+y^2=169$ and the point is $\left(-5,12\right)$ .

Calculation:

Compare the given equation of circle with the general equation of circle ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

This gives the center of the circle is $\left(0,0\right)$ and the radius of the circle is $13$ .

As the tangent line at the point $\left(-5,12\right)$ is perpendicular to the line joining center and point $\left(-5,12\right)$ as it is radius.

Use the formula for slope of radius.

  $\begin{array}{c}{m}_1=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{12-0}{-5-0}\\ =-\frac{12}{5}\end{array}$

Both the lines are perpendicular. So, $m_1{m}_2=-1$ .

  $\begin{array}{c}{m}_1{m}_2=-1\\ \left(-\frac{12}{5}\right)m_2=-1\\ m_2=\frac{5}{12}\end{array}$

So, the slope of the tangent line is $\frac{5}{12}$ .

Use the general equation of the tangent line.

  $\begin{array}{l}y=mx+b\\ y=\left(\frac{5}{12}\right)x+b\end{array}$

As the tangent line passes through the point $\left(-5,12\right)$ . So, substitute $-5$ for $x$ and $12$ for $y$ in the equation of tangent line.

  $\begin{array}{c}y=\left(\frac{5}{12}\right)x+b\\ 12=\left(\frac{5}{12}\right)\left(-5\right)+b\\ b=12+\frac{25}{12}\\ b=\frac{169}{12}\end{array}$

Now substitute $\frac{169}{12}$ for $b$ in the tangent line.

  $\begin{array}{c}y=\left(\frac{5}{12}\right)x+\frac{169}{12}\\ 12y=5x+169\\ 5x-12y+169=0\end{array}$

Therefore, The equation of the tangent line to the circle $x^2+y^2=169$ at the point $\left(-5,12\right)$ is $5x-12y+169=0$ .