Question

how to solve this further? ​

Answer 1

Answer:

x = √(1 - y²) or √(1 + y)(1 - y)

y = √(1 - x²) or √(1 + x)(1 - x)

Step-by-step explanation:

1 = √(x² + y²)

=> 1 = x² + y²    (Squaring both sides)

=> x² + y² - 1 = 0

So, y² = 1 - x²

∴ y = √(1 - x²) or √(1 + x)(1 - x)    (Using a² - b² = (a + b)(a - b) identity)

Similarly, x = √(1 - y²) or √(1 + y)(1 - y)

Answer 2

Given :  1 = √(x² + y²)

To Find : Possible values of x & y

Solution:

1 = √(x² + y²)

=> √(x² + y²)  = 1

Squaring both sides

=> x² + y² = 1

x² ,  y² ≥ 0  ( as square of real numbers can not be negative)

x² + y² = 1

=>  x² ,  y² ≤  1

0  ≤   x²  ≤  1

=>   -1 ≤   x   ≤  1

0  ≤   y²  ≤  1

=>   -1 ≤   y   ≤  1

x ∈ [-1, 1]

y  ∈ [-1, 1]

x² + y² = 1 is basically an equation of circle centered at origin ( 0 , 0) with radius 1

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5x(1+1/x^2+y^2)=12 ,5y(1-1/x^2+y^2)=4 solve it for x,y​ - Brainly.in

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