Intersecting circles using Newton's Method

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The discussion revolves around using Newton's Method to find the intersection of two circles defined by equations. The initial approach of setting the equations equal to each other led to a linear equation, which is not suitable for Newton's iteration. Clarification is provided that one should set the functions equal to each other rather than the constants. The importance of showing detailed work is emphasized to receive better guidance. The participant has yet to choose an initial condition for the iteration process.
Robb
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Homework Statement


upload_2017-11-8_20-33-41.png


Homework Equations

The Attempt at a Solution


My initial thought was to set the two equations equal to each other but the resulting equation is linear which gives a constant for a Newton iteration. I thought about Taylor's theorem in 2-d but I'm not so sure about that as far as deriving the iterating function. Please help!
 

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Robb said:

Homework Statement


View attachment 214649

Homework Equations

The Attempt at a Solution


My initial thought was to set the two equations equal to each other but the resulting equation is linear which gives a constant for a Newton iteration. I thought about Taylor's theorem in 2-d but I'm not so sure about that as far as deriving the iterating function. Please help!
Technically, you don't "set equations equal to each other." If you have one equation in the form f(x) = b and another equation in the form g(x) = b, then you can set f(x) equal to g(x). However, it makes no sense to write ##(x - 2)^2 + (y - 1)^2 = 2 = 3.5 = (x - 2.5)^2 + y^2##.

For the equations in your problem, what you said amounts to setting 2 = 3.5, which is obviouly untrue.

What was the work that you did? You need to show us what you did, rather than just loosely describe your work, so that we can steer you in the right direction.
 
upload_2017-11-9_6-16-48.png


I have not chosen an initial condition yet so I can get F(X).
 

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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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