Question

If $x$ and $y$ satisfy the relation $(x-1{)}^2+y^2=1$, then the possible value of $(x+y)$ is equal to

• A. $\frac{-3}{2}$
• B. $\frac{5}{2}$
• C. 3
• D. $\frac{-1}{4}$

Answer 1

Answer: (D)

$\text{ Let }x=1+\cos \theta \text{ and }y=\sin \theta$
$\therefore x+y=1+\cos \theta +\sin \theta$
$\text{ Hence }1-\sqrt{2}\le x+y\le 1+\sqrt{2}$