Finding intersections of tangents on circles

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Homework Help Overview

The discussion revolves around finding the intersection points of tangents to a circle centered at the origin with a radius of 2. The tangents have slopes of √3/3 and -√3/3, and participants are exploring how to calculate the intersection points of these tangents.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equation of the circle and how to derive it from the given information. There are attempts to differentiate the circle's equation and find the slopes of the tangents. Questions arise about the correctness of calculations and the implications of the tangent slopes.

Discussion Status

The discussion is active, with participants providing guidance on deriving the equation of the circle and calculating the slopes of the tangents. Some participants express confusion about the calculations and the implications of the results, while others offer clarifications and encourage further exploration of the problem.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is also an emphasis on understanding the symmetry of the circle and how it relates to the tangent lines.

  • #31
VanKwisH said:
apparently it's at o .... but isn't that impossibile?
Why would it be impossible? The y-value is approx 2.31 and the radius of the circle is only 2, so the tangents are just a little above the circle.
 
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  • #32
wtf i am completely lost now ... sorry for being a retard but how is the approx value 2.31
 
  • #33
mann so many different things and i can't seem to see wtf is going on O.o
 
  • #34
VanKwisH said:
wtf i am completely lost now ... sorry for being a retard but how is the approx value 2.31
The tangent lines intersect at x=0

y_1^{'}=\frac{-x}{\sqrt3}+\frac{4}{\sqrt3}

y_1^{'}=\frac{0}{\sqrt3}+\frac{4}{\sqrt3}

y_1^{'}=y_2^{'}=\frac{4}{\sqrt3}\approx 2.31
 
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  • #35
ooo ic u got it from 4/root 3 ... but i thought we canceled it so it was only
-x/root3 = x/root3
 
  • #36
VanKwisH said:
mann so many different things and i can't seem to see wtf is going on O.o
Don't have a graphing calculator?
 
  • #37
VanKwisH said:
ooo ic u got it from 4/root 3 ... but i thought we canceled it so it was only
-x/root3 = x/root3
B/c I was solving for the x value!
 
  • #38
yes i have a graphing calculator ...
 
  • #39
alright

so they point of intersection is at 0, and at 2.31
 
  • #40
VanKwisH said:
alright

so they point of intersection is at 0, and at 2.31
Yes.

I'm going to sleep now, you did good. If you're still not clear on some things, graph the circle and your 2 tangents and you'll see that everything you did was correct. Looks cool I guess.
 
  • #41
alright thanks man ur a fcukin genious
 
  • #42
rocophysics said:
Yes.

I'm going to sleep now, you did good. If you're still not clear on some things, graph the circle and your 2 tangents and you'll see that everything you did was correct. Looks cool I guess.

ahahah everything looks and sounds right ... too bad my paper looks all messed
up from all the erasing a scribbling but thnx for the explanations and not giving up on me :D
 
  • #43
hmmmmmm apparently ... there is 4 solutions and (0,2.31) is only one ...
i know another one should be (0,-2.31) but how do i solve for that and the other ones??
 
  • #44
Yes there is 4 solutions, if you checked it on the graph, you will notice that they will intersect at 4 locations.

Calculate all intersection or intersections of these tangents

I forgot all the questions you had on your original post.

For the first 2, we were given the slopes and we found the points at which the slopes lied. Well, now we use the fact that a circle is symmetrical with respects to the x and y axes. Can you find those points? By finding these points, you will know what your slope needs to be.
 

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