Chapter 8.1, Problem 54PS

Solution

To find: The diameter of a circular dish.

The original diameter of the dish is $\approx 11.75\text{in}\text{.}$ .

Given data: An archaeological dig is covered by a circular dish. The three points on the edge of dish are $\left(-4,2\right),\left(0,0\right)$ and $\left(6,4\right)$ . The measure of diameter of dish is required.

Method/Formula used:

The general equation of a circle is given by $x^2+y^2+2gx+2fy+c=0$ whose center lies at point $\left(-g,-f\right)$ and radius R of this circle is $R=\sqrt{g^2+f^2-c}$ .

Calculations:

The archaeological dig is covered by the circular dish. Let equation of the circle representing circumference/edge of the circular dish is

  $x^2+y^2+2gx+2fy+c=0$ … (1)

The three points on the edge of dish are $\left(-4,2\right),\left(0,0\right)$ and $\left(6,4\right)$ . These points must satisfy equation (1),

Substituting $-4$ for x and 2 for y in equation (1),

  $\left.\begin{array}{c}{\left(-4\right)}^2+{\left(2\right)}^2+2g\left(-4\right)+2f\left(2\right)+c=0\\ 16+4-8g+4f+c=0\\ -8g+4f+c=-20\\ 8g-4f-c=20\end{array}\right\}$ … (2)

Substituting 0 for x and 0 for y in equation (1),

  $\left.\begin{array}{c}{\left(0\right)}^2+{\left(0\right)}^2+2g\left(0\right)+2f\left(0\right)+c=0\\ c=0\end{array}\right\}$ … (3)

Substituting $-4$ for x , 2 for y and 0 for c in equation (1),

  $\left.\begin{array}{c}{\left(6\right)}^2+{\left(4\right)}^2+2g\left(6\right)+2f\left(4\right)+0=0\\ 36+16+12g+8f=0\\ 12g+8f=-52\\ 3g+2f=-13\end{array}\right\}$ … (4)

Substitute 0 for c in equation (2),

  $\left.\begin{array}{c}8g-4f-0=20\\ 4g-2f=10\end{array}\right\}$ … (5)

Adding equation (4) and (5),

  $\begin{array}{c}\left(3g+2f\right)+\left(4g-2f\right)=-13+10\\ 7g=-3\\ g=-3/7\end{array}$

Substitute $-3/7$ for g in equation (5),

  $\begin{array}{c}2g-f=5\\ f=2g-5\\ =2\left(-\frac{3}{7}\right)-5\\ f=-\frac{41}{7}\end{array}$

The radius of the circle is $R=\sqrt{g^2+f^2-c}$ , substitute $-\frac{3}{7}$ for g, $-\frac{41}{7}$ for f and 0 for c ,

  $\begin{array}{c}R=\sqrt{{\left(-\frac{3}{7}\right)}^2+{\left(-\frac{41}{7}\right)}^2-0}\\ =\sqrt{\frac{1690}{49}}\\ =5.87\end{array}$

Since each unit represents1inch, therefore, the diameter of the dish is $5.87\times 2\approx 11.75\text{in}\text{.}$ .