The equation of a circle ?

The typical equation of a circle is: (x - xo)² + (y - yo)² = R² → where:

xo: abscissa of center

yo: ordinate of center

R: radius of circle


The epicenter is 3 miles away from A (0 ; 0). So the epicenter belongs to the circle C₁.

The Center of the circle C₁ is A (0 ; 0) and the radius is 3.

The equation of the circle C₁ is:

(x - xA)² + (y - yA)² =  3²

(x - 0)² + (y - 0)² = 9

x² + y² = 9

x² = 9 - y² ← equation (1)


The epicenter is 3 miles away from B (3 ; 3). So the epicenter belongs to the circle C₂.

The Center of the circle C₂ is B (3 ; 3) and the radius is 3.

The equation of the circle C₂ is:

(x - xB)² + (y - yB)² =  3²

(x - 3)² + (y - 3)² = 9 → recall (1): x² + y² = 9

(x - 3)² + (y - 3)² = x² + y²

x² - 6x + 9 + y² - 6y + 9 = x² + y²

- 6x - 6y + 18 = 0

- x - y + 3 = 0

x = - y + 3 ← equation (2)

x² = (- y + 3)² → recall (1): x² = 9 - y²

(- y + 3)² = 9 - y²

y² - 6y + 9 = 9 - y²

2y² - 6y = 0

2y.(y - 3) = 0

y.(y - 3) = 0


Firt case: y = 0 → recall (2): x = - y + 3

x = 3

Point M (3 ; 0)


Second case: (y - 3) = 0 → y = 3 → recall (2): x = - y + 3

x = 0

Point N (0 ; 3)


You can see that you obtain 2 points M and N.

You must check if the point N belongs to the circle C₃ or if the point M belongs to the circle C₃.


The epicenter is 3 miles away from C (- 3 ; 3). So the epicenter belongs to the circle C₃.

The Center of the circle C₃ is C (- 3 ; 3) and the radius is 3.

The equation of the circle C₃ is:

(x - xC)² + (y - yC)² =  3²

(x + 3)² + (y - 3)² =  3²


First case: does the point M (3 ; 0) belong on to the circle C₃ ?

(x + 3)² + (y - 3)² =  3² → when: x = 3, then: y = 0

(3 + 3)² + (0 - 3)² =  3²

36 + 9 =  9 ← no possible, so the point M don't belong to the circle C₃.


First case: does the point N (0 ; 3) belong on to the circle C₃ ?

(x + 3)² + (y - 3)² =  3² → when: x = 0, then: y = 3

(0 + 3)² + (3 - 3)² =  3²

9 + 0 = 9 ← ok, so the point N don't belong to the circle C₃.


As the point N (0 ; 3) belongs to the circle circle C₁ and belongs to the circle C₂ and belongs to the circle C₃, you can deduce that this point N (0 ; 3) is the solution.

Set 1 unit = 1 km on the coordination plane.

Let (a, b) be the epicenter of the Black Box.

Distance between (a, b) and (0, 0) = 3
√[(a - 0)² + (b - 0)²] = 3
a² + b² = 9 …… [1]

Distance between (a, b) and (3, 3) = 3
√[(a - 3)² + (b - 3)²] = 3
a² - 6a + 9 + b² - 6a + 9 = 9
a² + b² - 6a - 6b = -9 …… [2]

Distance between (a, b) and (-3, 3) = 3
√[(a + 3)² + (b - 3)²] = 3
a² + 6a + 9 + b² - 6a + 9 = 9
a² + b² + 6a - 6b = -9 …… [3]

[1] - [2]:
6a + 6b = 18
a + b = 3 …… [4]

[1] - [3]:
-6a + 6b = 18
-a + b = 3 …… [5]

[4] - [5]:
2a = 0
a = 0

[4] + [5]:
2b = 6
b = 3

Hence, epicenter = (0, 3)

(x - h)^2 + (y - k)^2 = r^2
(0 - h)^2 + (0 - k)^2 = 3^2
(3 - h)^2 + (3 - k)^2 = 3^2
(-3 - h)^2 + (3 - k)^2 = 3^2

h^2 + k^2 = 9
(3 - h)^2 = 9 - (3 - k)^2
(-3 - h)^2 = 9 - (3 - k)^2

(3 - h)^2 = (-3 - h)^2
9 - 6h + h^2 = 9 + 6h + h^2
-6h = 6h
0 = 12h
0 = h

h^2 + k^2 = 9
0^2 + k^2 = 9
k^2 = 9
k = -3 , 3

(3 - h)^2 + (3 - k)^2 = h^2 + k^2
(3 - 0)^2 + (3 - k)^2 = 0^2 + k^2
9 + 9 - 6k + k^2 = k^2
18 - 6k = 0
18 = 6k
3 = k

(x - 0)^2 + (y - 3)^2 = 3^2

(h , k) is the center

(0 , 3)

The 3 points A,B & C is equidistant from the epicenter of the black box. This implies it is a center of a circle with the radius
=3 miles.
Let the equation of the circle be
(x-h)^2+(y-k)^2=9
inserting the coordinates of the 3 points, get
h^2+k^2=9----(1)
(3-h)^2+(3-k)^2=9------(2)
(3+h)^2+(3-k)^2=9-----(3)
(2)-(3)=>
-12h=0=>
h=0
From (1), get k^2=9=>k=+/-3.
If k=-3, then both (2) & (3) does not hold=>k=3 miles.
So the epicenter of the black box is (0,3).
8) The center of vibration.

In other word, we have to find center Coordinates of a circle passing through, 

A (0,0) , B (3,3) and C (-3,3)

Gradient of AB (m1) = (3-0)/ (3-0)  =  + 1 and  that of AC (m2)  = (3-0)/(-3+0) = -1

Hence m1 * m2 = 1 * (-1) = - 1

Hence the two lines AB and AC are at right angles to each other.also  A, B and C are situated on the circle, 

This is possible iff ( if and only if ) BC is diameter. 

Mid point of this circle is  the required epicenter. Let its  = coordinates be ( h , k )

h = ( 3 - 3 )/2    = 0 and  k   =  (3+3)/2  = 3

Hence Epicenter is at ( 0, 3 )  .................... Answer

  

The epicenter of the box is located at intersection of the right bisectors of AB
and BC. Slope AB = (3-0)/(3-0) = 1, midpoint AB = [(3/2),(3/2)], eqn of right
bisector of AB is [y-(3/2)]/[x-(3/2)] = -1, ie., y-(3/2) = (3/2)-x, ie., x+y = 3...(1).
Slope BC = (3-3)(-6) = 0, midpoint BC = (0,3), eqn of right bisector of BC is
x = 0...(2). Solving (1) & (2) gives (x,y) = (0,3), the epicenter of the Black Box.