Solving for an x^2+y^2=Ae^x Circle Passing Through (0,1)

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The equation x^2 + y^2 = Ae^x describes a curve, not a circle, which leads to confusion in the discussion. To find a circle that passes through the point (0,1) and is perpendicular to the curve, the user suggests using the square root of y to derive the slope of the tangent line. This slope is then applied to the linear equation y = ax + b to determine the equation of the desired circle. The conversation highlights the importance of clarifying the geometric properties of the given equation. Ultimately, the user successfully finds the equation of the tangent line.
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I had x^2+y^2=Ae^x, how can i find out the cirle pass (0,1) and perpendicular to that cirle ? :frown:
 
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Phyphy said:
I had x^2+y^2=Ae^x, how can i find out the cirle pass (0,1) and perpendicular to that cirle ? :frown:

Perpendicular to what circle? x2+ y2= Aex is NOT a circle!
 
sorry, it is curve not circle ? But I foun answer, I do sqare root of y and find y' and let it be a tangent's coeficent, and put it in a of y=ax+b, i found equation of y. :smile: :smile:
 

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