Chapter 13.7, Problem 35WE

a.

To determine

To find : the center of the circle

Answer to Problem 35WE

The center of the circle is $\left(2,0\right)$

Explanation of Solution

Given information :

Passes through $\left(2,10\right),\left(10,6\right),\text{ and }\left(-6,-6\right)$

Formula used :

  $\begin{array}{l}\text{the slope of the line }=\frac{y-y_1}{x-x_1}\text{ and }\\ \text{the point }x\text{ of the center }=\frac{m_1\cdot m_2\left(y_3-y_1\right)+m_1\left(x_2+x_3\right)-m_2\left(x_1+x_2\right)}{2\left(m_1-m_2\right)}\text{ and }\\ \text{the point }y\text{ of the center }=\frac{-1}{m_1}\left(x-\frac{x_1+x_2}{2}\right)+\frac{y_1+y_2}{2}\end{array}$

Calculation :

Let $\left(x,y\right)$ be the point of the circle

Let $\left(x_1{y}_1\right),\left(x_2,y_2\right)\text{ and }\left(x_3,y_3\right)$ be the point of $\left(2,10\right),\left(10,6\right)\text{ and }\left(-6,-6\right)$

The slope of $m_1$

  $\begin{array}{l}=\frac{10-6}{2-10}\\ =\frac{4}{-8}\\ =-\frac{1}{2}\end{array}$

The slope of $m_2$

  $\begin{array}{l}=\frac{6-\left(-6\right)}{10-\left(-6\right)}\\ =\frac{6+6}{10+6}\\ =\frac{12}{16}\\ =\frac{3}{4}\end{array}$

To find the point $x$ of the center

  $\begin{array}{l}=\frac{\frac{-1}{2}\cdot \frac{3}{4}\left(-6-10\right)+\frac{-1}{2}\left(10+\left(-6\right)\right)-\frac{3}{4}\left(2+10\right)}{2\left(\frac{-1}{2}-\frac{3}{4}\right)}\\ =\frac{\frac{-3}{8}\left(-16\right)+\frac{-1}{2}\cdot 4-\frac{3}{4}\cdot 12}{2\left(\frac{-2-3}{4}\right)}\\ =\frac{6-2-9}{\frac{-5}{2}}\\ =\frac{-5\cdot 2}{-5}\\ =2\end{array}$

To find the point $y$ of the center

  $\begin{array}{l}=\frac{-1}{\frac{-1}{2}}\left(2-\frac{2+10}{2}\right)+\frac{10+6}{2}\\ =2\left(\frac{4-12}{2}\right)+8\\ =2\cdot -4+8\\ =-8+8\\ =0\end{array}$

Hence, the center of the circle is $\left(2,0\right)$ .