Chapter 1 - Graphs - 1.4 Circles - 1.4 Assess Your Understanding - Page 40: 57

$(1,0)$

First, we have to find the equation of the line perpendicular to the tangent line $x-2y+4=0$ at $(0, 2)$. Note that the slopes of perpedicular lines are negative reciprocals of each other. Write $x-2y+4-0$ in slope-intercept form ($y=mx+b$) then identify its slope: $x-2y+4-0$ $-2y=-x-4$ $y=\frac{x}{2}+2$ The slope of the tangent line is $\frac{1}{2}$. This means that the slope of the line perpendicular to it is $-2$, which means the tentative equation of the line is $y=-2x+b$. Since the line passes through $(0, 2)$, substitute $x=0$ and $y=2$ to find the value of $b$: $y=-2x+b$ $2=-2(0)+b$ $b=2$ Therefore, the equation of the line perpendicular to the tangent line is $y=-2x+2$. Next, find the equation of the line perpendicular to the tangetn line $y=2x-7$ at $(3, -1)$ The slope of the tanget line is $2$ so the slope of the line perpendicular to it is $-\frac{1}{2}$. This means that the tentative equation of the perpendicular line is $y=-\frac{1}{2}x+b$. Since the line passes through $(3, -10$, substitute $x=3$ and $y=-1$ to the tentative equation to obtain: $y=-\frac{1}{2}x+b$ $-1=-\frac{1}{2}(3)+b$ $-1=-\frac{3}{2}+b$ $-1+\frac{3}{2}=b$ $\frac{1}{2}=b$ Thus, the equation of the perpendicular line is $y=-\frac{1}{2}x+\frac{1}{2}$. The lines $y=-2x+2$ and $y=-\frac{1}{2}x+\frac{1}{2}$ both pass through the center of the circle. The center of the circle is their point of intersection. To find the point of intersection, equate the two equations to each other to obtain: $\begin{align*}-2x+2&=-\frac{1}{2}x+\frac{1}{2}\\ -2x+\frac{1}{2}x&=\frac{1}{2}-2\\ -\frac{3}{2}x&=-\frac{3}{2}\\ x&=1\end{align*}$ SOlve for $y$ by substituting $x=1$ to $y=-2x+2$ to obtain: $y=-2x+2\\ y=-2(1)+2\\ y=0$ Thus, the point of intersetion is $(1, 0)$. This is also the center of the circle.