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.

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.

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

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$$\begin{gathered} y=-2x+b,this\tan gentlinepassesthrough(0,2)thenb=2,then \\ y=-2x+2,thisistheperpendiuclarequationofx-2y+4=0. \end{gathered}$$

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 berole="math" localid="1646301625126" $$\begin{gathered} y=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.x+b,andthispas\sin gthrough(3,-1),thenb=\frac{1}{2},then \\ y=\frac{-1}{2}x+\frac{1}{2} \\ 2y=-x+1,thisisequationofperpendicularlineofy=2x-7 \end{gathered}$$

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 ,

$$\begin{gathered} y=-2x+2\Rightarrow y+2x=2\to \left(1\right) \\ 2y=-x+1\Rightarrow 2y+x=1\to \left(2\right) \\ \left(1\right)\times 2\Rightarrow 2y+4x=4\to \left(3\right) \\ \left(3\right)-\left(2\right)wegetx=1,substitutein\left(1\right)theny=0 \\ (x,y)=(1,0)isthecenterofthecircle. \end{gathered}$$