Explaining the Yes/No Response in Two Points on a Circle

  • Thread starter Thread starter blimkie
  • Start date Start date
  • Tags Tags
    Circle Points
AI Thread Summary
The discussion revolves around understanding why both points A and B on a circle yield a "yes" response in a given scenario. It clarifies that the y-components of the vectors OA and OB will cancel each other out, making the length of OC dependent solely on the x-components. For OC to lie on the circle, the sum of the x-components must equal the radius, while it will lie inside the circle if this sum is less than the radius. The angles formed by the vectors with the x-axis significantly influence the outcome. Overall, the explanation emphasizes the geometric relationships between the points and their components.
blimkie
Messages
111
Reaction score
0
http://putfile.com/pic.php?pic=10/30314175469.jpg&s=x11

if anyone would like to explain this to me that would be excelent and most appreciated

i know that a and b are both "yes" but i i don't know why or how to explain it

thanks
kyle
 
Last edited by a moderator:
Physics news on Phys.org
How, exactly, do you know both a and b are "yes?"
 
wouldnt point c be on the circle if points a and b were separated a certain distance
 
Lets call OC the x-axis and the y-axis is perpendicular to it. since the y components will always cancel out, the length of OC will be the sum of the X components of each vector (OA,OB). For OC to be on the circle its length has to be equal to the radius, so OC will be on the circle if the sum of OA's and OB's x components equal r (radius), for example if they each make a 60 deg. angle with the x-axis (.5r + .5r = r). For OC to lie inside the circle its length has to be less than r, this can happen if the sum of the x components is less than r, for example, OA and OB each make a more than 60 deg. angle with the x axis.
 
thanks a lot daniel_i_l that explains it better than i could
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top