Chapter 8.7, Problem 44PS

Solution

a.

Find the equations of the circle centered at Lajitas, Mexico and La paz.

Lajitas equation form $x^2+y^2=49$ .

Mexico City equation form ${(x-3)}^2+{(y+7)}^2=9$ .

La Paz equation form ${(x+4)}^2+{(y+4)}^2=25$ .

Given:

Relative position of the seismic stations are described below.

Mexico city $700$

  $miles$ south and $300$

  $miles$ east of Lajitas

La paz $400$

  $miles$ south and $400$

  $miles$ west of Lajitas.

Calculation:

The goal of this task is to write an equation for each circle. In order to do so going to use the formula for a circle with a center at $(h,k)$ and radius $r$ that looks like

  ${(x-h)}^2+{(y-k)}^2=r^2$

First, will write an equation for Lajitas Since it is located at origin $h=0$ and $k=0$ , also radius is 700 miles but because 100 miles $=1$ unit radius $r=7$ and equation looks like

  $x^2+y^2=49$

After that equation for Mexico City.

Since it is $300miles$ to East $h=3$ and $700miles$ South $k=-7$ , also radius is $300miles$ so $r=3$ , now substitute that in standard form for circle.

  ${(x-3)}^2+{(y+7)}^2=9$

Lastly, let's write an equation for La Paz.

Since it is $400miles$ to West $h=-4$ and 400 miles South $k=-4$ , also radius is $500miles$ so $r=5$ , now substitute that in standard form for circle so get.

  ${(x+4)}^2+{(y+4)}^2=25$

Conclusion:

Lajitas equation form $x^2+y^2=49$ .

Mexico City equation form ${(x-3)}^2+{(y+7)}^2=9$ .

La Paz equation form ${(x+4)}^2+{(y+4)}^2=25$ .

b.

New Equations for lajitas

The New equations are

  $8x+8y+56=0$

and

  $-6x+14y+98=0$ .

Given:

Relative position of the seismic stations are described below.

Mexico city $700$

  $miles$ south and $300$

  $miles$ east of Lajitas

La paz $400$

  $miles$ south and $400$

  $miles$ west of Lajitas.

Calculation:

The circles for Lajitas, La Paz, and Mexico City.

  $\begin{array}{c}{x}^2+y^2=7^2\\ $ {(x+4)}^2+{(y+4)}^2=5^2$$ {(x-3)}^2+{(y+7)}^2=3^2$\end{array}$

Multiply out the 2nd and 3rd equations, then $\text{ substitute }{7}^2=49\text{ for }\left(x^2+y^2\right)\text{. }$

  $\begin{array}{c}$ {(x+4)}^2+{(y+4)}^2=25$x^2+8x+16+y^2+8y+16=25\\ \left(x^2+y^2\right)+8x+8y+7=0\\ 49+8x+8y+7=0\\ 8x+8y+56=0\cdots \cdots \cdots \cdots (1)\\ \\ $ {(x-3)}^2+{(y+7)}^2=3^2$x^2-6x+9+y^2+14y+49=9\\ \left(x^2+y^2\right)-6x+14y+49=0\\ 49-6x+14y+49=0\\ -6x+14y+98=0\cdots \cdots \cdots \cdots (2)\end{array}$

Conclusion:

The New equations are $8x+8y+56=0$ and $-6x+14y+98=0$ .

c.

Co ordinates of the epicenter for solving system of linier equations.

Now since each unit is $100miles$ know that epicenter was at $\text{(0,-7)(0,-7) }$ or $\text{(0,-700)(0,-700)}$

Given:

Relative position of the seismic stations are described below.

Mexico city $700$

  $miles$ south and $300$

  $miles$ east of Lajitas

La paz $400$

  $miles$ south and $400$

  $miles$ west of Lajitas.

Calculation:

In order to find the epicenter, need to solve the system of two linear equations. are given equations.

  $\begin{array}{r}\hfill -6x+14y+98=0\\ \hfill 8x+8x+56=0\end{array}$

From second equation can express $y=-x-7$

Now substitute the value of $y$ in the first equation to get the value of $x$

  $\begin{array}{c}-6x+14(-x-7)+98=0\cdots \cdots \cdots \cdots (3)\\ -6x-14x-98+98=0\cdots \cdots \cdots \cdots (4)\\ -20x=0\\ x=0\cdots \cdots \cdots \cdots (5)\end{array}$

Now substitute $x$ back into first equation.

  $\begin{array}{c}8x+8y+56=0\cdots \cdots \cdots \cdots (6)\\ 0+8y=-56\cdots \cdots \cdots \cdots (7)\\ y=-7\cdots \cdots \cdots \cdots (8)\end{array}$

Now since each unit is $100miles$ know that epicenter was at $\text{(0,-7)(0,-7) }$ or $\text{(0,-700)(0,-700)}$

Conclusion:

Now since each unit is $100miles$ know that epicenter was at $\text{(0,-7)(0,-7) }$ or $\text{(0,-700)(0,-700)}$