Geometry problem

[NotAllThere]

Given a continuous, differentiable curve, find the mid point and radius of a circle that bisects the y-axis and is tangential to the curve at a specific point on the curve.

It looks something like the blue circle in this diagram. These are Mohr's circles.

We know the point where the circle touches the curve, we know the gradient at that point, we know that the circle must be perpendicular to the x axis at y = 0. I think that's sufficient information. But how to calculate it? Geometrically or, more interesting, analytically?

 

[_V_]

 

[OwlHoot]

How is the curve represented?

* Explicitly, via an equation of the form y = f(x), e.g. y = e^x - 5

* Implicitly, via f(x, y) = 0, e.g. x^3 y^2 + cos(x) log(y) = 0, from where one cannot necessarily derive an explicit finite expression for either x or y

* Parametrically, via x = u(t), y = v(t), e.g. x = e^(-2t), y = e^(-3t)

 

[BlasterBates]

The answer to your question is here:

The equation of the tangent to a circle - Equations of circles - Higher - CCEA - GCSE Maths Revision - CCEA - BBC Bitesize

 

[TestMangler]

Scooter says you should take a short position on that stock immediately !

 

[OwlHoot]

The answer to your question is here:

The equation of the tangent to a circle - Equations of circles - Higher - CCEA - GCSE Maths Revision - CCEA - BBC Bitesize

It isn't, because in this problem the circle centre is constrained to be on the y-axis, which isn't generally the case for an osculating circle.

 

[BlasterBates]

It isn't, because in this problem the circle centre is constrained to be on the y-axis, which isn't generally the case for an osculating circle.

Yes but you simply set y=0 in whatever equation you figure out. You have the tangent, now you find the circle with y=0.

If you didn't have the circle constrained you would have an infinite number of solutions.

 

[AtW]

Who's got a geometry problem?

 

[northernladuk]

 

[courtg9000]

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