Chapter 9.1, Problem 107E

To determine

To find: The equation of the tangent line to the circle ${\left(x+2\right)}^2+{\left(y-1\right)}^2=100$ at the point $\left(4,9\right)$ .

Answer to Problem 107E

The equation of the tangent line to the circle ${\left(x+2\right)}^2+{\left(y-1\right)}^2=100$ at the point $\left(4,9\right)$ is $3x+4y-48=0$ .

Explanation of Solution

Given information:

The equation of the circle is ${\left(x+2\right)}^2+{\left(y-1\right)}^2=100$ and the point is $\left(4,9\right)$ .

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 $\left(-2,1\right)$ and the radius of the circle is $10$ .

As the tangent line at the point $\left(4,9\right)$ is perpendicular to the line joining center and point $\left(4,9\right)$ as it is radius.

Use the formula for slope of radius.

  $\begin{array}{c}{m}_1=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{9-1}{4-\left(-2\right)}\\ =\frac{8}{6}\\ =\frac{4}{3}\end{array}$

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

  $\begin{array}{c}{m}_1{m}_2=-1\\ \left(\frac{4}{3}\right)m_2=-1\\ m_2=-\frac{3}{4}\end{array}$

So, the slope of the tangent line is $-\frac{3}{4}$ .

Use the general equation of the tangent line.

  $\begin{array}{l}y=mx+b\\ y=\left(-\frac{3}{4}\right)x+b\end{array}$

As the tangent line passes through the point $\left(4,9\right)$ . So, substitute $4$ for $x$ and $9$ for $y$ in the equation of tangent line.

  $\begin{array}{c}y=\left(-\frac{3}{4}\right)x+b\\ 9=\left(-\frac{3}{4}\right)4+b\\ 9=-3+b\\ b=12\end{array}$

Now substitute $12$ for $b$ in the tangent line.

  $\begin{array}{c}y=\left(-\frac{3}{4}\right)x+12\\ 4y=-3x+48\\ 3x+4y-48=0\end{array}$

Therefore, The equation of the tangent line to the circle ${\left(x+2\right)}^2+{\left(y-1\right)}^2=100$ at the point $\left(4,9\right)$ is $3x+4y-48=0$ .