- #1
Phyphy
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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 ? 
Solving for an x^2+y^2=Ae^x Circle Passing Through (0,1)
- Thread starter Phyphy
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SUMMARY
The discussion centers on the equation x² + y² = Ae^x, which is identified as a curve rather than a circle. The user seeks to find a circle that passes through the point (0,1) and is perpendicular to the curve. The solution involves taking the square root of y to derive y', which represents the slope of the tangent line. This slope is then used to formulate the equation of the tangent line in the form y = ax + b.
PREREQUISITES- Understanding of differential calculus, specifically derivatives
- Familiarity with the concept of tangent lines to curves
- Knowledge of implicit differentiation
- Basic algebra for manipulating equations
- Study implicit differentiation techniques for curves
- Learn about the properties of tangent lines and their equations
- Explore the relationship between curves and circles in coordinate geometry
- Investigate the applications of exponential functions in geometry
Students and educators in mathematics, particularly those focusing on calculus and geometry, as well as anyone interested in the properties of curves and their tangents.
- #2
HallsofIvy
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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 ?
Perpendicular to what circle? x2+ y2= Aex is NOT a circle!
- #3
Phyphy
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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. 