Chapter 9.1, Problem 103E

To determine

To find: The equation of the tangent line to the circle $x^2+y^2=25$ at the point $\left(3,-4\right)$ .

Answer to Problem 103E

The equation of the tangent line to the circle $x^2+y^2=25$ at the point $\left(3,-4\right)$ is $4y-3x+25=0$ .

Explanation of Solution

Given information:

The equation of the circle is $x^2+y^2=25$ and the point is $\left(3,-4\right)$ .

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 $5$ .

As the tangent line at the point $\left(3,-4\right)$ is perpendicular to the line joining center and point $\left(3,-4\right)$ as it is radius.

Use the formula for slope of radius.

  $\begin{array}{c}{m}_1=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{-4-0}{3-0}\\ =-\frac{4}{3}\end{array}$

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 $\left(3,-4\right)$ . So, substitute $3$ for $x$ and $-4$ for $y$ in the equation of tangent line.

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

Now substitute $-\frac{25}{4}$ for $b$ in the tangent line.

  $\begin{array}{c}y=\left(\frac{3}{4}\right)x-\frac{25}{4}\\ 4y=3x-25\\ 4y-3x+25=0\end{array}$

Therefore, The equation of the tangent line to the circle $x^2+y^2=25$ at the point $\left(3,-4\right)$ is $4y-3x+25=0$ .