Latest self Aptitude Question SOLUTION: If x^2 +y^2-4x-4y+8=0, then the value of x-y is

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(#M40010615) SELF question Keep an EYE

If x^2 +y^2-4x-4y+8=0, then the value of x-y is

Arjun (11 years ago)

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SDeep Bhattacharya  (11 years ago)

@ SAUMITRA DAY :

U have made a little mistake.

In ur solution we can find
(x-2)^2 + (y-2)^2=0

now, if x,y are real quantities
then only we can write now if the sum of 2 or more than 2 square terms of real quantities is zero then the terms are individually zero.

So,(x-2)^2=0 and (y-2)^2=0
x=2
y=2,so x-y=0..

So,(x-y)=0.

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Garima  (11 years ago)

x^2 +y^2-4x-4y+8=0
x^2 -4x+4 +y^2-4y+4=0
(x-2)^2 + (y-2)^2=0

I think there is some mistake signs of some terms in qn.

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Saumitra Dey  (11 years ago)

x^2 +y^2-4x-4y+8=0
x^2 -4x+4 +y^2-4y+4=0
(x-2)^2 + (y-2)^2=0
now if the sum of 2 or more than 2 is zero then this are individually zero.
now (x-2)^2=0 and (y-2)^2=0
so x = +2 or -2 and y = +2 or -2
so the value of x-y = 0 or +4 or -4

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