Chapter 0.1, Problem 51E

To determine

To find: Find the equation of line that is tangent to the circle at point (3,4).

Answer to Problem 51E

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 origin. It passes through the point $\left(3,4\right)$ .

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 $\left(0,0\right)$ and $\left(3,4\right)$ .

Substitute $\left(x_1=0,y_1=0,x_2=3.y_2=4\right)$ .

Slope of radius is:

  $\begin{array}{l}{m}_1=\frac{4-0}{3-0}\\ m_1=\frac{4}{3}\end{array}$

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:

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

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

  $\begin{array}{l}y-4=\frac{-3}{4}\left(x-3\right)\\ 4\left(y-4\right)=-3\left(x-3\right)\\ 4y-16=-3x+9\\ 4y=-3x+9+16\\ 4y=-3x+25\end{array}$

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