Question

If $x,y\in R$, satisfy the equation $x^2+y^2-4x-2y+5=0$, then the value of the expression $\frac{(\sqrt{x}-\sqrt{y}{)}^2+4\sqrt{xy}}{x+\sqrt{xy}}$, is

• A. $\sqrt{2}+1$
• B. $\frac{\sqrt{2}+1}{2}$
• C. $\frac{\sqrt{2}-1}{2}$
• D. $\frac{\sqrt{2}+1}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{\sqrt{2}+1}{\sqrt{2}}$

Given equation is

$x^2+y^2-4x-2y+5=0\Rightarrow {(x-2)}^2+{(y-1)}^2=0$
$\Rightarrow x=2,y=1$ ...(1)

Now,
$\frac{({(\sqrt{x}-\sqrt{y})}^2+4\sqrt{xy})}{(x+\sqrt{xy})}=\frac{{(\sqrt{x}+\sqrt{y})}^2}{(x+\sqrt{xy})}=\frac{{(\sqrt{x}+\sqrt{y})}^2}{\sqrt{x}(\sqrt{x}+\sqrt{y})}=\frac{(\sqrt{x}+\sqrt{y})}{\sqrt{x}}$

Substituting values from (1) , we get
$\frac{({(\sqrt{x}-\sqrt{y})}^2+4\sqrt{xy})}{(x+\sqrt{xy})}=\frac{(\sqrt{2}+\sqrt{1})}{\sqrt{2}}=\frac{(\sqrt{2}+1)}{\sqrt{2}}$
Hence, option 'D' is correct.