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September 30th, 2012, 03:22 PM   #1
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finding a point on a circle given the slope of the tan line

x^(2) + y^(2) = 100


mT = -3/4


Can someone show me how to go about finding this?
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September 30th, 2012, 03:48 PM   #2
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Re: finding a point on a circle given the slope of the tan l

You can transform circle equation into function for upper and lower semicircle:

and now take derivative
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September 30th, 2012, 04:02 PM   #3
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Re: finding a point on a circle given the slope of the tan l

forgive me but wouldn't I take the Derv to find the slope of the tan line?

I am trying to find a point on the upper half of the circle.

I have tried to think on my own hence: rise/run and point slope y-y1=m(x-x1)

when I to that I get (3/4, 0)


would this be right?
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September 30th, 2012, 04:43 PM   #4
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Re: finding a point on a circle given the slope of the tan l

Hello, mathkid!

Quote:
















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September 30th, 2012, 05:01 PM   #5
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Re: finding a point on a circle given the slope of the tan l

Wow!
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September 30th, 2012, 08:46 PM   #6
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Re: finding a point on a circle given the slope of the tan l

Another approach:



Implicitly differentiate with respect to x:





Equate this to



Substitute into original equation:





Take positive root.



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October 1st, 2012, 06:49 PM   #7
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Re: finding a point on a circle given the slope of the tan l

You don't have to differentiate at all. Any tangent line to a circle is perpendicular to the radius at that point. If the slope of the tangent line is -3/4, then the radius at that point has slope 4/3. But if (x,y) is a point on the circle, then the slope of the line from the center, the origin (0, 0), is y/x. That is, y/x= 4/3 and, of course, . From y/x= 4/3, y= (4/3)x so or , and then .
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