Chapter 9.1, Problem 105E

To determine

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

Answer to Problem 105E

The equation of the tangent line to the circle $x^2+y^2=12$ at the point $\left(2,-2\sqrt{2}\right)$ is $2y-\sqrt{2}x+6\sqrt{2}=0$ .

Explanation of Solution

Given information:

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

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 $2\sqrt{3}$ .

As the tangent line at the point $\left(2,-2\sqrt{2}\right)$ is perpendicular to the line joining center and point $\left(2,-2\sqrt{2}\right)$ as it is radius.

Use the formula for slope of radius.

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

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

  $\begin{array}{c}{m}_1{m}_2=-1\\ \left(-\sqrt{2}\right)m_2=-1\\ m_2=\frac{\sqrt{2}}{2}\end{array}$

So, the slope of the tangent line is $\frac{\sqrt{2}}{2}$ .

Use the general equation of the tangent line.

  $\begin{array}{l}y=mx+b\\ y=\left(\frac{\sqrt{2}}{2}\right)x+b\end{array}$

As the tangent line passes through the point $\left(2,-2\sqrt{2}\right)$ . So, substitute $2$ for $x$ and $-2\sqrt{2}$ for $y$ in the equation of tangent line.

  $\begin{array}{c}y=\left(\frac{\sqrt{2}}{2}\right)x+b\\ -2\sqrt{2}=\left(\frac{\sqrt{2}}{2}\right)\left(2\right)+b\\ b=-2\sqrt{2}-\sqrt{2}\\ b=-3\sqrt{2}\end{array}$

Now substitute $-3\sqrt{2}$ for $b$ in the tangent line.

  $\begin{array}{c}y=\left(\frac{\sqrt{2}}{2}\right)x-3\sqrt{2}\\ 2y=\sqrt{2}x-6\sqrt{2}\\ 2y-\sqrt{2}x+6\sqrt{2}=0\end{array}$

Therefore, The equation of the tangent line to the circle $x^2+y^2=12$ at the point $\left(2,-2\sqrt{2}\right)$ is $2y-\sqrt{2}x+6\sqrt{2}=0$ .