Question

in the graph below, points p, q, and r are on the circle, and $overline{pq}$ is the diameter of the circle. a. what is the length of the diameter of the circle? b. what are the coordinates of the center of the circle? c. what is an equation for the perpendicular bisector of $overline{qr}$?

Explanation:

Step1: Find diameter length using distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For points $P(-1,3)$ and $Q(2,-4)$, we have $x_1=-1,y_1 = 3,x_2=2,y_2=-4$. Then $d=\sqrt{(2-(-1))^2+(-4 - 3)^2}=\sqrt{(3)^2+(-7)^2}=\sqrt{9 + 49}=\sqrt{58}$.

Step2: Find center - mid - point of diameter

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For $P(-1,3)$ and $Q(2,-4)$, the center $O$ has coordinates $(\frac{-1+2}{2},\frac{3+( - 4)}{2})=(\frac{1}{2},-\frac{1}{2})$.

Step3: Find perpendicular bisector of $QR$

First, find the mid - point of $QR$. For $Q(2,-4)$ and $R(-3,-2)$, the mid - point $M$ is $(\frac{2+( - 3)}{2},\frac{-4+( - 2)}{2})=(-\frac{1}{2},-3)$.
The slope of $QR$ is $m_{QR}=\frac{-2-( - 4)}{-3 - 2}=\frac{2}{-5}=-\frac{2}{5}$. The slope of the perpendicular bisector $m$ is the negative reciprocal of $-\frac{2}{5}$, so $m=\frac{5}{2}$.
Using the point - slope form $y - y_1=m(x - x_1)$ with point $M(-\frac{1}{2},-3)$ and $m = \frac{5}{2}$, we have $y+3=\frac{5}{2}(x+\frac{1}{2})$. Expanding gives $y+3=\frac{5}{2}x+\frac{5}{4}$, or $y=\frac{5}{2}x+\frac{5}{4}-3=\frac{5}{2}x-\frac{7}{4}$.

Answer:

A. $\sqrt{58}$
B. $(\frac{1}{2},-\frac{1}{2})$
C. $y=\frac{5}{2}x-\frac{7}{4}$