Known $x\ge0,x^2+(y-2)^2=1,W=\dfrac{3x^2+2\sqrt{3}xy+5y^2}{x^2+y^2 }$, find the maximum value of $W$.
Hint:
When $x\ne0$, set $t=\dfrac{y}{x}$ from the graph to know $t\ge\sqrt{3},W=5+\dfrac{2\sqrt{3}t- 2}{1+t^2}\in(5,6]$
When $x=0$, obviously $W=5$, so $W\in[5,6]$