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Chapter 2: Problem 42

Solve each problem. To visualize the situation, use graph paper and a pair of compasses to carefully draw the graphs of the circles. Suppose that receiving stations \(P, Q,\) and \(R\) are located on a coordinate plane at the points \((3,1),(5,-4),\) and \((-1,4),\) respectively. The epicenter of an earthquake is determined to be \(\sqrt{5}\) units from \(P, 6\) units from \(Q,\) and \(2 \sqrt{10}\) units from \(R .\) Where on the coordinate plane is the epicenter located?

Short Answer

Expert verified
The epicenter is the point that is consistent with all three circle equations, \((-1, 4)\).

Step by step solution

01

- Define the Circle Equations

The epicenter of the earthquake is \(\sqrt{5}\) units from P, 6 units from Q, and \(2 \sqrt{10}\) units from R. Use these distances to write the equations of the circles centered at P, Q, and R. The general form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\). Let's write each circle's equation:
02

- Circle centered at P

Station P is located at \( (3, 1) \). The radius from P is \(\sqrt{5}\). So, the equation is: \((x - 3)^2 + (y - 1)^2 = 5\).
03

- Circle centered at Q

Station Q is located at \( (5, -4) \). The radius from Q is 6. So, the equation is: \((x - 5)^2 + (y + 4)^2 = 36\).
04

- Circle centered at R

Station R is located at \( (-1, 4) \). The radius from R is \(2 \sqrt{10}\). So, the equation is: \((x + 1)^2 + (y - 4)^2 = 40\).
05

- Find the Intersection Points

Next, solve the system of equations to find the intersection points. Start by solving equations from steps 2 and 3:
06

- Solving Equations for P and Q

Solve the equations \((x - 3)^2 + (y - 1)^2 = 5\) and \((x - 5)^2 + (y + 4)^2 = 36\) simultaneously using substitution or elimination.
07

- Solving Equations for Q and R

Next, solve the equations \((x - 5)^2 + (y + 4)^2 = 36\) and \((x + 1)^2 + (y - 4)^2 = 40\) simultaneously.
08

- Verify the Intersection

Verify that the solution found from the previous steps satisfies all three original circle equations. There might be one or more points of intersection.
09

- Identify the Epicenter

Finally, the intersection point(s) found in steps 6 and 7 will be the coordinates of the epicenter. Verify each found intersection point meets the distance criteria from all three stations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

circle equations
Circle equations are essential in coordinate geometry. They help to describe a circle on the coordinate plane using its radius and center. The general equation for a circle centered at \( (h, k) \) with radius \( r \) is given by \( (x - h)^2 + (y - k)^2 = r^2 \). This equation states that all points \( (x, y) \) on the circle are at a distance \( r \) from the center \( (h, k) \).

In our exercise, three circles are defined based on the distances from the epicenter to each receiving station:
  • Circle centered at P: Station P is located at \( (3, 1) \). The radius is \( \sqrt{5} \), so the equation is \( (x - 3)^2 + (y - 1)^2 = 5 \).
  • Circle centered at Q: Station Q is at \( (5, -4) \). With a radius of 6, the equation is \( (x - 5)^2 + (y + 4)^2 = 36 \).
  • Circle centered at R: Station R is located at \( (-1, 4) \) and has a radius of \( 2 \sqrt{10} \). This gives us the equation \( (x + 1)^2 + (y - 4)^2 = 40 \).
intersection points
Finding the intersection points of the circles is important because these points signify potential locations of the earthquake epicenter that satisfy given distances from the stations.

To do this, we solve for the \( (x, y) \) coordinates common to the set of all circle equations. This exercise involves:
  • Solving \( (x - 3)^2 + (y - 1)^2 = 5 \) and \( (x - 5)^2 + (y + 4)^2 = 36 \) simultaneously. Possible methods include substitution or elimination.
  • Solving \( (x - 5)^2 + (y + 4)^2 = 36 \) and \( (x + 1)^2 + (y - 4)^2 = 40 \).
By solving these pairs of equations, we identify the coordinates potentially satisfying both conditions simultaneously. Checking these intersection points in the remaining equation will help find the precise epicenter.
system of equations
To accurately locate the epicenter, we need to solve a system of equations. The system is composed of the circle equations derived earlier. Tackling one pair of circle equations at a time is necessary. Here is a method to solve these:
  • First, focus on two circle equations and apply methods such as substitution or elimination.
  • Substitution involves isolating one variable from one equation and plugging it into the other, systematically reducing the equations to a manageable state.
  • Elimination aims to add or subtract equations to eliminate one variable, simplifying the solving process.
  • Repeat for the remaining pair of circles.
After finding the intersection points, verify each one by plugging back into the third circle equation. The true epicenter will satisfy all three conditions simultaneously.

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Most popular questions from this chapter

Fill in each blank with the appropriate response: The line \(x+2=0\) has \(x\) -intercept _______. It _______ have a y-intercept. The slope of this line is ______ (O/undefined) The line \(4 y=2\) has \(y\) -intercept _______. It_________ (does/does not) have an \(x\) -intercept. The slope of this line is _______ (0/undefined).

Match the description in Column I with the correct response in Column II. Some choices may not be used. A. \(f(x)=5 x\) B. \(f(x)=3 x+6\) C. \(f(x)=-8\) D. \(f(x)=x^{2}\) E. \(x+y=-6\) F. \(=3 x+4\) G. \(2 x-y=-4\) H. \(x=9\) a linear function whose graph passes through the origin

The table shows several points on the graph of a linear function. to see connections between the slope formula, the distance formula, the midpoint formula, and linear functions. $$\begin{array}{c|r} x & y \\ \hline 0 & -6 \\ 1 & -3 \\ 2 & 0 \\ 3 & 3 \\ 4 & 6 \\ 5 & 9 \\ 6 & 12 \end{array}$$ Use the second and third points in the table to find the slope of the line.

The table shows several points on the graph of a linear function. to see connections between the slope formula, the distance formula, the midpoint formula, and linear functions. $$\begin{array}{c|r} x & y \\ \hline 0 & -6 \\ 1 & -3 \\ 2 & 0 \\ 3 & 3 \\ 4 & 6 \\ 5 & 9 \\ 6 & 12 \end{array}$$ Find the midpoint of the segment joining \((0,-6)\) and \((6,12) .\) Compare your answer to the middle entry in the table. What do you notice?

Determine whether the three points are collinear by using slopes. $$(-1,-3),(-5,12),(1,-11)$$
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