If $x$ and $y$ are both positive and $\sqrt{x^2+y^2}=3x-y,$ then $\dfrac{x}{y} =$

If $x$ and $y$ are both positive and $\sqrt{x^2+y^2}=3x-y,$ then $\dfrac{x}{y} =$

A. $0$
B. $\dfrac14$
C. $\dfrac12$
D. $\dfrac34$
E. $\dfrac32$

Answer: D

Source: Magoosh

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