Chapter 8, Problem 3TP

Solution

To calculate: The point $(55,25)$ located at inside or outside the crater.

The point is located at inside the crater.

Given information: Three points on the edge of the circular crater represented by the coordinates $(-35,35),(0,0),(40,80)$

Calculation:

Let $\left(x_0,y_0\right)$ be the center and $r>0$ the radius of the edge circle of the Copernicus crater. Then the circle is given by the equation

  ${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2=r^2$

The points $(-35,35),(0,0),(40,80)$ lie on the circle, so their coordinates satisfy the equation of the circle. Substituting them for $x$ and $y$ , the result will be like the following $3\times 3$ system:

  ${\left(-35-x_0\right)}^2+{\left(35-y_0\right)}^2=r^2$

  ${\left(0-x_0\right)}^2+{\left(0-y_0\right)}^2=r^2$

  ${\left(40-x_0\right)}^2+{\left(80-y_0\right)}^2=r^2$

Applying the binomial formula, the system can be written in the following way:

  $\begin{array}{c}1225+70x_0+x_0^2+1225-70y_0+y_0^2=r^2\\ x_0^2+y_0^2=r^2\\ 1600-80x_0+x_0^2+6400-160y_0+y_0^2=r^2 \end{array}$

Subtracting the second equation from the first one

  $\begin{array}{c}1225+70x_0+1225-70y_0=0\\ 70x_0-70y_0=-2500\\ 7x_0-7y_0=-250\end{array}$

Subtracting the second equation from the third one

  $\begin{array}{c}1600-80x_0+6400-160y_0=0\\ -80x_0-160y_0=-8000\\ x_0+2y_0=100 \end{array}$

Now have a $2\times 2$ linear system:

  $\begin{array}{c}7x_0-7y_0=-250\\ x_0+2y_0=100\end{array}$

From the second equation

  $x_0=100-2y_0$

Substituting this into the first one,

  $\begin{array}{c}7\left(100-2y_0\right)-7y_0=-250\\ 700-14y_0-7y_0=-250\\ -21y_0=-250-700\\ -21y_0=-950\\ y_0=\frac{950}{21}\end{array}$

Substituting this into the expression for $x_0$ ,

  $\begin{array}{c}{x}_0=100-2\cdot \frac{950}{21}\\ =\frac{2100-2\cdot 950}{21}\\ =\frac{200}{21}\end{array}$

The centre is $\frac{950}{21},\frac{200}{21}$ and the radius is

  $\begin{array}{c}r=\sqrt{x_0^2+y_0^2}\\ =\sqrt{{\frac{950}{21}}^2+{\frac{200}{21}}^2}\\ =\frac{\sqrt{942500}}{21}\\ =\frac{50\sqrt{377}}{21}\end{array}$

The point $(x,y)=(55,25)$ will be inside the crater if and only if it satisfies

  ${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2\le r^2$

That means that the distance of $(x,y)$ from the center $\left(x_0,y_0\right)$ is less than $r$ . Then calculate:

  ${\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2=55-{\frac{950}{21}}^2+25-\frac{{200}^2}{21}$

  $\begin{array}{c}={\frac{205}{21}}^2+{\frac{325}{21}}^2\\ =\frac{147650}{{21}^2} \end{array}$

Since $147650<942500$ , let’s conclude that the point $(55,25)$ lies inside the crater.