\(x^2+y^2=100\). All of the following could be true EXCEPT

(A) |x|+|y|=10

(B) |x|>|y|

(C) |x|>|y|+10

(D) |x|=|y|

(E) |x|−|y|=5

PS: Did my best to search for an existing post and couldn't find any.

A. |x|+|y|=10 is possible if one is 0 and the other is 10.

B. |x|>|y| is possible if |x|>|52√| and |y|<|52√|

C. |x|>|y|+10 is never possible because if |x|>10, x2+y2 becomes greater than 100, which is wrong.

D. |x|=|y| is possible if each is equal to |52√|.

E. |x|−|y|=5 is possible if |x|=|9.11| and |y|=|4.11|.

Therefore all but C are possible. |x|>|y|+10 means x is greater than 10, which is not possible

Is there any alternate way to tackle this problem besides just plugging in numbers? It's very challenging to figure out 9.11 and 4.11 could be two possibilities to rule out E.

Here's what I did to solve this problem. Got stuck at this step and did not have a clue to proceed further...Any algebraic help will get kudos

\(x^2+y^2=100\)

\((x+y)^2-2xy = 10^2\)

\((x+y)^2 = \frac{10^2}{-2xy}\)

\(|x +y| = \frac{|10|}{\sqrt{-2xy}}\) => Root of negative number will result in imaginary and therefore it's likely that this condition might be the answer. But I'm not even sure if this is the correct approach and could not relate |x+y| and |10|.

Merging similar topics. Please refer to the solutions on page 1 and ask if anything remains unclear.