sigma_108
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Part 'a' is straightforward however I am confused as to how to tackle part 'b'. One method that does work is to draw a new circle centre (-5,-9), radius 5 and then find the minimum argument using the distance from centre to origin to form a triangle and then incorporate arcsin into the expression. I don't actually fully understand why this works - I just though about it as a translation of the locus and even then it's a hazy explanation.

I would appreciate it if somebody could explain why my method works and also a different, more efficient solution (if there is one). Thanks
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laurawatt
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Hi, could you post the question please?
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sigma_108
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(Original post by laurawatt)
Hi, could you post the question please?
Sorry - completely forgot to attach the image! Thanks.
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laurawatt
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(Original post by sigma_108)
Sorry - completely forgot to attach the image! Thanks.
Hi, I’m sorry, I don’t know how to answer part b)
Hopefully this bumps the thread so that someone else can help!
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sigma_108
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(Original post by laurawatt)
Hi, I’m sorry, I don’t know how to answer part b)
Hopefully this bumps the thread so that someone else can help!
No worries. Thanks
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the bear
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so if you have z - 11 - 10i that is like a vector from 11,10 to z.

the locus of z is the circle from part a. so we need to consider lines from the point (11,10) to the circle. they want to know which line has the most negative angle, measured from the horizontal direction to the right at point B in the picture.
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NotNotBatman
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it becomes obvious when you draw a sketch.

arg(z-11-10i) is a half line extending from 11+10i to z , which is tangential to your circle since you want the minimum. mess around with some triangles and the answer pops out.
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