Question

If $x,y\in R$ satisfy the equation $x^2+y^2-4x-2y+5=0$ then the value of the expression
$\frac{\left[\right(\sqrt{x}-\sqrt{y}{)}^2+4\sqrt{xy})]}{(x+\sqrt{xy})}$ 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 $\frac{\sqrt{2}+1}{\sqrt{2}}$

$x^2+y^2-4x-2y+5=0$

$x^2-4x+4+y^2-2y+1=0$

$(x-2{)}^2+(y-1{)}^2=0$

$x=2$ and $y=1$ Since $(x-2{)}^2=0$ and $(y-1{)}^2=0$

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

$=\frac{(\sqrt{2}-\sqrt{1}{)}^2+4\sqrt{2}}{2+\sqrt{2}}$

$=\frac{3-2\sqrt{2}+4\sqrt{2}}{2+\sqrt{2}}=\frac{3+2\sqrt{2}}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}$

$=\frac{6-4+4\sqrt{2}-3\sqrt{2}}{4-2}$

$=\frac{2+\sqrt{2}}{2}=\frac{\sqrt{2}+1}{\sqrt{2}}$