Q.55 - Precalculus Enhanced with Graphing Utilities

Question

Refer to Problem 52. The line x - 2y + 4 = 0 is tangent to a circle at (0, 2). The line 

 y=2x- 7 is tangent to the same

circle at (3, -1). Find the center of the circle.

Step-by-Step Solution

Answer

The center of the circle is (1,0)

Step 1.Given information

Here the line x - 2y + 4 = 0 is tangent to a circle at (0, 2). The line y=2x- 7 is tangent to the same circle at (3, -1)  is given.

We have to find out the center of the cirle.

Step 2. Description of find out the equation of perpendicular of x-2y+4=0

Here the line is $x-2y+4=0$, first convert this equation to slope-intercept form, we get $$\begin{gathered} x-2y+4=0 \\ -2y=-x-4 \sin ce slope intercept form is y=mx+b \\ y=\frac{x}{2}+2 \end{gathered}$$

 then the slope of this tangent is $\frac{1}{2}$so the line is perpendicular  to it , thus the

slope = -2.Hence equation  will becomes $$\begin{gathered} y =-2x+b ,this \tan gent line passes through(0,2) then b=2, then \\ y=-2x+2 , this is the perpendiuclar equation of x-2y+4=0. \end{gathered}$$

 

Step 3. Description of finding the perpendicular equation of y = 2 x - 7

Here the line is already in the slope-intercept form, $y=2x-7$and the slope is 2. So line perpendicular to it has slope $-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$., then the equation will be  $$\begin{gathered} y=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.x+b , and this pas\sin g through (3,-1) , then b=\frac{1}{2}, then \\ y=\frac{-1}{2}x+\frac{1}{2} \\ 2y=-x+1 , this is equation of perpendicular line of y=2x-7 \end{gathered}$$

Step 4. Description of finding the center of the circle

So here two lines that intersect at the center of the circle.  $$\begin{gathered} y=-2x+2 \\ 2y=-x+1 \end{gathered}$$

  now  to get the center of the circle we have to solve this equations ,

$$\begin{gathered} y=-2x+2 \Rightarrow y+2x=2 \to \left(1\right) \\ 2y=-x+1 \Rightarrow 2y+x=1 \to \left(2\right) \\ \left(1\right)\times 2 \Rightarrow 2y +4x=4 \to \left(3\right) \\ \left(3\right)-\left(2\right) we get x=1 , substitute in \left(1\right) then y=0 \\ (x,y)=(1,0) is the center of the circle. \end{gathered}$$