Question

Suppose that receiving stations​ X, Y, and Z are located on a coordinate plane at the points ​(4​,5​), ​(-6​,-6​), and ​(-14​,2​), respectively. The epicenter of an earthquake is determined to be 10 units from X​, 5 units from​ Y, and 13 units from Z. Where on the coordinate plane is the epicenter​ located?
Find the coordinates of the epicenter.

Answer 1

Answer:

  (-2, -3)

Step-by-step explanation:

A careful graph shows the point (-2, -3) is at the intersection of the circles whose radii are the given distances from the receiving stations.

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The simultaneous equations for the circles can be solved algebraically.

The epicenter is 10 units from X, so lies on the circle ...

  (x -4)^2 +(y -5)^2 = 10^2

  x^2 -8x +16 +y^2 -10y +25 = 100

  x^2 +y^2 -8x -10y = 59

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The epicenter is 5 units from Y, so lies on the circle ...

  (x +6)^2 +(y +6)^2 = 5^2

  x^2 +12x +36 +y^2 +12y +36 = 25

  x^2 +y^2 +12x +12y = -47

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The epicenter is 13 units from Z, so lies on the circle ...

  (x +14)^2 +(y -2)^2 = 13^2

  x^2 +28x +196 +y^2 -4y +4 = 169

  x^2 +y^2 +28x -4y = -31

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Subtracting the second equation from each of the other two, we get ...

  (x^2 +y^2 -8x -10y) -(x^2 +y^2 +12x +12y) = (59) -(-47)

  -20x -22y = 106 . . . . eq1 -eq2

  (x^2 +y^2 +28x -4y) -(x^2 +y^2 +12x +12y) = (-31) -(-47)

  16x -16y = 16 . . . . . . . .eq3 -eq2

These simultaneous linear equations can be solved a variety of ways. We might use substitution:

  x = y+1 . . . . . from eq3 -eq2 divided by 16

  10(y +1) +11y = -53 . . . . . from eq1 -eq2 divided by -2

  21y = -63 . . . . . . . . . . . . simplify, subtract 10

  y = -3

  x = y+1 = -2

The epicenter is located at (x, y) = (-2, -3).