Chapter 0.1, Problem 57E

To determine

To find: Find the equation of line that is tangent to the circle at point $\left(-2,6\right)$ .

Answer to Problem 57E

The equation of tangent line is $4y=-3x+25$ .

Explanation of Solution

Given information:

It is given that the radius of the circle is $5$ and center is $\left(1,2\right)$ .

Concept used: Tangent line to the circle and radius of circle are perpendicular to each other.

Calculation:

The tangent line to the circle and the radius of the circle are perpendicular to each other.

The slope of line that passes through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is:

  $m=\frac{y_2-y_1}{x_2-x_1}$

Radius passes through the points $\left(1,2\right)$ and $\left(-2,6\right)$ .

Substitute $\left(x_1=1,y_1=2,x_2=-2.y_2=6\right)$ .

Slope of radius is:

  $\begin{array}{l}{m}_1=\frac{6-2}{-2-1}\\ m_1=\frac{4}{-3}\end{array}$

If two lines are perpendicular then product of their slope is $-1$ that is $m_1\cdot m_2=-1$

Let slope of the tangent line is $m$ then $m_1\cdot m=-1$

Substitute $m_1=\frac{-4}{3}$ .

Slope of tangent line is:

  $\begin{array}{l}-\frac{4}{3}\cdot m=-1\\ -4m=-3\\ m=\frac{-3}{-4}\\ m=\frac{3}{4}\end{array}$

Now the equation of tangent line which passes through the point $\left(x_1,y_1\right)$ and having slope $m$ is:

  $y-y_1=m\left(x-x_1\right)$

Substitute $x_1=-2,y_1=6,m=\frac{3}{4}$ .

  $\begin{array}{l}y-6=\frac{3}{4}\left(x-\left(-2\right)\right)\\ 4\left(y-6\right)=3\left(x+2\right)\\ 4y-24=3x+6\\ 4y=3x+6+24\\ 4y=3x+30\end{array}$

The equation of tangent line is $4y=3x+30$ .