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

[Phyphy]

I had x^2+y^2=Ae^x, how can i find out the cirle pass (0,1) and perpendicular to that cirle ?

 

[HallsofIvy]

I had x^2+y^2=Ae^x, how can i find out the cirle pass (0,1) and perpendicular to that cirle ?

Perpendicular to what circle? x²+ y²= Ae^x is NOT a circle!

 

[Phyphy]

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.