Finding intersections of tangents on circles

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SUMMARY

The discussion revolves around calculating the intersection points of tangents to a circle centered at the origin with a radius of 2. The equation of the circle is established as x² + y² = 4. Two tangents with slopes of √3/3 and -√3/3 are analyzed, leading to the identification of intersection points at (0, 2.31) and (0, -2.31). The participants emphasize the importance of using the point-slope form and implicit differentiation to derive the necessary equations for solving the problem.

PREREQUISITES
  • Understanding of circle equations, specifically x² + y² = r²
  • Knowledge of implicit differentiation techniques
  • Familiarity with the point-slope form of a linear equation
  • Ability to solve systems of equations for intersection points
NEXT STEPS
  • Study implicit differentiation in calculus to enhance problem-solving skills
  • Learn about the point-slope form and its applications in finding tangent lines
  • Explore graphing techniques for visualizing circles and their tangents
  • Investigate symmetry in geometric figures to predict intersection points
USEFUL FOR

Students studying calculus, particularly those focusing on geometry and tangent lines, as well as educators seeking to clarify concepts related to circle equations and differentiation.

  • #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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