Question : If $x^{2}+y^{2}+6x+5=4(x-y)$, then $(x-y)$ is:
Option 1: $1$
Option 2: $0$
Option 3: $–1$
Option 4: $4$
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Correct Answer: $1$
Solution : Given: $x^{2}+y^{2}+6x+5=4(x-y)$ ⇒ $x^{2}+y^{2}+6x+5-4x+4y=0$ ⇒ $x^{2}+y^{2}+2x+5+4y=0$ ⇒ $x^{2}+2x+1+y^{2}+4y+4=0$ ⇒ $x^{2}+2×x×1+1+y^{2}+2×y×2+4=0$ ⇒ $(x+1)^{2}+(y+2)^{2}=0$ ⇒ $x$ = – 1 and $y$ = – 2 [If the sum of two square terms is zero, then each term will also be zero] So, $(x-y) = (–1–( –2))=1$ Hence, the correct answer is $1$.
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