# Why is $$y$$ squared?

Calculus Level 4

Consider the differential equation $\frac{dy}{dx}=\frac{\sqrt{x^{2}+y^{2}}-x}{y}$

Given that $$y(0)=4$$, if the equation $$y(x)=0$$ has exactly one negative (real) root $$\beta$$, find the value of $$\beta$$.

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