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A circle rests in the interior of the parabola with an equation y=x^2 so that it is tangent to the parabola at two points. How much higher is the center of the circle than the points of tangency?


Thank you very much!


 Dec 29, 2018

Is there any more info given for this question.......like the size or equation for the circle?  I think there are MANY circles that would fit the bill.....do you need an answer in terms of the circle size/equation?

 Dec 29, 2018

Sorry for answering late...


CPhill already got the answer. If you could look at his, you may understand :)

CoolStuffYT  Jan 1, 2019

Because of symmetry, the center of the circle will lie  at  (0, k)


Let the equation of the  circle be


x^2 + (y - k)^2 = r^2

x^2 + y^2 - 2ky + k^2 = r^2

Using implicit differentiation, the slope of any tangent line to the circle is given by


2x + 2yy' - 2ky'  = 0

x + yy'  - ky' = 0

y'  ( y - k)  =  -x

y' =     [ x ] / [k - y]


The slope of the tangent line to the parabola at any point =   d/dx (x^2) =     2x


So....equating these slopes, we have that


2x =  x / [ k - y]

2 = 1 / [ k - y]

k - y =    1/2

y = k -1/2      this is the y coordinate of the intersection of the parabola and the circle


So.....the difference in the height of the circle's center and the tangent point(s) is just the difference in their  y values




k - (k - 1/2)  =    1/2   unit



cool cool cool

 Dec 29, 2018
edited by CPhill  Dec 29, 2018

Great! It took me a couple minutes to understand, but once I got your method it worked out! Thanks!

CoolStuffYT  Dec 30, 2018

OK, CS.....!!!!


[ BTW....this is kind of a neat problem..... ]



cool cool cool

CPhill  Dec 30, 2018

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