Chapter 1.4, Problem 55AYU

To determine

To find: the center of the circle.

Answer to Problem 55AYU

Therefore, the coordinates of the center $\left(x,y\right)$ is $\left(1,0\right)$ .

Explanation of Solution

Given:

The line $x-2y+4=0$ is tangent to a circle at $\left(0,2\right)$ .

The line $y=2x-7$ is tangent to the same circle at $\left(3,-1\right)$ .

Since the line containing the center and the tangent line are perpendicular, the product of their slopes will be $-1$ .

Let $m$ denote the slope of the line containing the center, $m_1$ denote the slope of the tangent line.

  $x-2y+4=0$ , and $m_2$ denote the slope of the tangent line $y=2x-7$ .

Consider the tangent line $x-2y+4=0$ at the point $\left(0,2\right)$ .

Rearrange the equation to the form $y=mx+c,$ where $m$ is the slope of the line.

Subtract $x+4$ from both sides of the equation.

  $\begin{array}{l}x-2y+4-x-4=0-x-4\\ -2y=-\left(x+4\right)\end{array}$

Divide both side of the equation by $-2$ .

  $\begin{array}{l}\frac{-2y}{-2}=\frac{-\left(x+4\right)}{-2}\\ y=\frac{1}{2}\left(x+4\right)\end{array}$

Therefore, the slope $m_1$ is $\frac{1}{2}$ .

The product of $m$ and $m_1$ is $-1$ .

  $\begin{array}{l}m{m}_1=-1\\ m.\frac{1}{2}=-1\\ m=-2\end{array}$

So, the slope $m$ is $-2$ .

Let $\left(x,y\right)$ denotes the coordinates of the center.

So, a point-slope form of the equation of the line containing the center can be written using

  $\left(y-y_1\right)=m_2\left(x-x_1\right)$

  $\begin{array}{l}y-2=-2\left(x-0\right)\\ y-2=-2x\\ 2x+y=2\end{array}$

Consider the tangent line $y=2x-7$ at the point $\left(3,-1\right)$ . This equation of the line is of the form $y=mx+c$ , where $m$ is the slope of the line.

Comparing the equations, get the slope of the tangent line as $2$ .

The product of $m$ and $m_2$ is $-1$ .

  $\begin{array}{l}m{m}_2=-1\\ m.2=-1\\ m=-\frac{1}{2}\end{array}$

Therefore, the slope $m_2$ is $-\frac{1}{2}$ .

Let $\left(x,y\right)$ denotes the coordinated of the center.

So, a point-slope form of the equation of the containing the center can be written using

  $\begin{array}{l}\left(y-y_1\right)=m_2\left(x-x_1\right)\\ y+1=-\frac{1}{2}\left(x-3\right)\\ 2y+2=-x+3\\ x+2y=1\end{array}$

Solve the equations for the line containing the center to get the coordinates of the center.

From the equation $x+2y=1$ , find the value of $x$ .

  $x=1-2y$

Substitute $1-2y$ for $x$ in the equation $2x+y=2$ , to solve by $y$ .

  $2\left(1-2y\right)+y=2$

Remove the parentheses using the distributive law and simplify.

  $\begin{array}{l}2-4y+y=2\\ 2-3y=2\end{array}$

Subtract $2$ from both the sides.

  $\begin{array}{l}2-3y-2=2-2\\ 3y=0\end{array}$

Therefore, $y=0$

Substitute the value of $y$ in any of the two equations. Consider the equation $x+2y=1$ .

  $\begin{array}{l}x+2\left(0\right)=1\\ x=1\end{array}$

Therefore, the coordinates of the center $\left(x,y\right)$ is $\left(1,0\right)$ .

Conclusion:

Therefore, the coordinates of the center $\left(x,y\right)$ is $\left(1,0\right)$ .