Finding intersections of tangents on circles

  • Thread starter VanKwisH
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In summary, the homework statement is trying to find the intersection of the two tangents of a circle with a radius of 2. The equation of the circle is x^2+y^2=4 and the slope of the first tangent is \frac{1}{\sqrt{3}} and the slope of the second tangent is -\frac{1}{\sqrt{3}}. The point of intersection is found by setting the two tangents equal to each other and solving for x.
  • #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?
 
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  • #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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