Chapter 16.1, Problem 9PS

To determine

To find: The value of $k$ given in the equation.

Answer to Problem 9PS

The value of $k$ given in the equation is $-1\text{or}-21$

Explanation of Solution

Given information:

The given equation of the line is,

  $3x-4y+k=0$

The centre of the circle is at $\left(5,1\right)$ and the radius of the circle is $2\text{unit}$ .

The radius of the circle and the distance from the centre to the graph of the given equation of the line are equal.

The value of $k$ given in the equationis calculated as,

  $\begin{array}{l}d=\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}\\ 2=\frac{\left|3\left(5\right)+\left(-4\right)\left(1\right)+k\right|}{\sqrt{{\left(3\right)}^2+{\left(-4\right)}^2}}\\ 10=\left|k+11\right|\\ k=-1\text{or}-21\end{array}$

Thus, the value of $k$ given in the equation is $-1\text{or}-21$ .