Find the value of x,y if the distances of the point (x,y) from (-3,0) as well as from (3,0) are 4

Given:  distances of the point (x,y) from (-3,0) as well as from (3,0) are 4

To find: $x,y$

Solution:

The distance between (x, y) from (-3, 0) is √[(x + 3)² + y²]   

 √[(x + 3)² + y²] = 4

[(x + 3)² + y²] = 16

x² + y² + 6x  - 7 = 0  ... (1)

The distance between (x, y) from (3, 0) is √[(x - 3)² + y²]    

√[(x - 3)² + y²] = 4

[(x - 3)² + y²] = 16

x² + y² - 6x  - 7 = 0  ... (2)

From (1) and (2) we get

6x  - 7 = - 6x  - 7

x = 0

Put x = 0 in (1) we get y = √7

So x=0 ,  y = √7

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Updated on: 10-Oct-2022

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