A particle moves along the circle $$x^2+y^2+4x-8y=108.$$ Determine it's position at the instant when the rate of change of $x$-coordinate is equal to the rate of change of its $y$-coordinate.
2026-02-24 03:47:11.1771904831
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Hint. Note that by differentiating $x^2+y^2+4x-8y=108$, we obtain $$2xx'+2y y'+4x'-8y'=0$$ Now you know that $x'=y'$. Can you take it from here?