# A point (x, y) is 4 m from (12, 6) and 8 m from (18, 4). What is the position of the point.

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### 1 Answer

The distance between two points `(x_1, y_1)` and `(x_2, y_2)` is `D = sqrt((y_2 - y_1)^2 + (x_2 - x_1)^2)` .

As the point X(x, y) is a distance of 4 m from (12, 6)

4 = sqrt((y - 6)^2 + (x - 12)^2)

=> 16 = (y - 6)^2 + (x - 12)^2

This is a circle with center (12, 6) and radius 4.

Similarly, the fact that X is at a distance 8 m from (18, 4) gives the equation 64 = (y - 4)^2 + (x - 18)^2

The point of intersection of the two circles 16 = (y - 6)^2 + (x - 12)^2 and 64 = (y - 4)^2 + (x - 18)^2 gives the coordinates of the point X.

16 = (y - 6)^2 + (x - 12)^2

=> 16 = y^2 - 12y + x^2 - 24x + 180 ...(1)

64 = (y - 4)^2 + (x - 18)^2

=> 64 = y^2 - 8y + x^2 - 36x + 340 ...(2)

(2) - (1)

=> 48 = 4y - 12x + 160

=> y = 3x - 28

Substitute this in 16 = (y - 6)^2 + (x - 12)^2

=> 16 = (3x - 34)^2 + (x -12)^2

The solution of this equation is `(57 - sqrt 39)/5` and `(57 + sqrt 39)/5`

The corresponding values of y are 2.453 and 9.9469

**The position of point X is either (10.151, 2.453) or (12.6489, 9.9469)**