Chapter 8.1, Problem 6GP

Solution

To find: The diameter of a circle.

The diameter of the circle is $20$ units.

Given data: Three points $\left(0,0\right)$ , $\left(6,-2\right),$ and $\left(16,8\right)$ lie on the circle.

Method/Formula used:

The perpendicular bisectors of two chords of a circle meet at the center of the circle. The equation of line with slope m and passing through point $\left(h,k\right)$ is $y-k=m\left(x-h\right)$ . The slope m of a line passing through two points $\left(x_1,y_1\right),\left(x_2,y_2\right)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$ . Two lines with slopes $m_1$ and $m_2$ are perpendicular when $m_1\cdot m_2=-1$ .

Calculations:

Three points $\left(0,0\right)$ , $\left(6,-2\right),$ and $\left(16,8\right)$ lie on the circle. The locations of points $O\left(0,0\right)$ , $A\left(6,-2\right),$

  $B\left(16,8\right)$ are shown in Fig. 1.

  

Lines OA and AB are two chords of the circle and M and N are the mid points of chords OA and AB respectively. Their bisectors MC and NC meet at point C , which is the center of the circle passing through points $O\left(0,0\right)$ , $A\left(6,-2\right),$

  $B\left(16,8\right)$ .

The midpoint of chord OA is $\left(\frac{0+6}{2},\frac{0-2}{2}\right)$ or $\left(3,-1\right)$ . The slope of line OA is $m_1=\frac{-2-0}{6-0}$ or $m_1=-\frac{1}{3}$ , therefore, slope of line MC, $m_2$ is given by

  $\begin{array}{c}{m}_2=-\frac{1}{m_1}\\ =-\frac{1}{-1/3}\\ =3\end{array}$

Therefore, equation of line MC , with slope 3 and passing through point $\left(3,-1\right)$

  $\left.\begin{array}{c}y-\left(-1\right)=3\left(x-3\right)\\ y=3x-10\end{array}\right\}$ … (1)

The midpoint of chord AB is $\left(\frac{16+6}{2},\frac{8-2}{2}\right)$ or $\left(11,3\right)$ . The slope of line AB is

  $\begin{array}{c}{m^{\prime }}_1=\frac{8-\left(-2\right)}{16-6}\\ =\frac{10}{10}\\ =1\end{array}$

Therefore, slope of line NC, ${m^{\prime }}_2$ is given by

  $\begin{array}{c}{m^{\prime }}_2=-\frac{1}{{m^{\prime }}_1}\\ =-\frac{1}{1}\\ =-1\end{array}$

Therefore, equation of line NC , with slope -1 and passing through point $\left(11,3\right)$

  $\left.\begin{array}{c}y-\left(3\right)=-1\left(x-11\right)\\ y-3=-x+11\\ y=-x+14\end{array}\right\}$ … (2)

Now, solve equation (1) and (2), equating y as,

  $\begin{array}{c}3x-10=-x+14\\ 4x=24\\ x=6\end{array}$

Substitute, -2 for x in equation (2),

  $\begin{array}{l}y=-x+14\\ y=-6+14\\ y=8\end{array}$

Therefore, the center of the circle is $C\left(6,8\right)$ . The CO is radius of the circle given by

  $\begin{array}{c}CO=\sqrt{{\left(6-0\right)}^2+{\left(8-0\right)}^2}\\ =\sqrt{36+64}\\ =10\end{array}$

Therefore, the diameter of the circle is $2\cdot CO$ , i.e., $20$ units.