# Homework Help: Finding intersections of tangents on circles

1. Jan 16, 2008

### VanKwisH

1. The problem statement, all variables and given/known data

A circle, with a center at the origin, and a radius of 2 has at least one tangent
with a slope of root3/3 and at least one tangent with a slope of -root3/3 .
Calculate all intersection or intersections of these tangents

2. Relevant equations

3. The attempt at a solution

how do i even do this?!?!?

2. Jan 16, 2008

### rocomath

What is the equation of your circle?

3. Jan 16, 2008

### VanKwisH

i don't have one .....that's the problem.... that is the exact question word for word ....

4. Jan 16, 2008

### rocomath

Yes you do, think back to how you can write the equation of the circle by just knowing it's position is at the Origin and that it's radius is 2.

5. Jan 16, 2008

### VanKwisH

(0 - h)^2 + (0 - k)^2 = 4 ??

6. Jan 16, 2008

### rocomath

Good try, but h & k = 0, not x & y.

$$x^2+y^2=4$$

So, take the derivative ... and solve for y'.

7. Jan 16, 2008

### VanKwisH

alright so 2x + 2y = 0
and 2y = -2x
then y = -2x / 2
y = -x ........... so wtf does that have to do with finding the intersect

8. Jan 16, 2008

### rocomath

You didn't even do it right so don't 'wtf', do it again and get y' if you still want help.

"Implicit differentiation"

9. Jan 16, 2008

### VanKwisH

AHHHHHHHHHH alrighty 1 sec lemme calculate this

10. Jan 16, 2008

### VanKwisH

okay

i got dy/dx = -2x/2y ......... then what??

11. Jan 16, 2008

### rocomath

$$y'=\frac{-x}{y}$$

We're told one of our slopes is

Tangent 1 = $$\frac {1}{\sqrt{3}} =\frac {-x}{y}$$ and also gives us our coordinate $$(-1,\sqrt{3})$$

Now just use the Point-slope equation for this one and the other one whose slope can obviously be found the same way.

And to find where the slopes intersect, set them equal to each other and solve for the x position.

Last edited: Jan 16, 2008
12. Jan 16, 2008

### VanKwisH

um ...... i thought the tangents slopes were root3 / 3 and negative root3 / 3

13. Jan 16, 2008

### VanKwisH

so then would root3/3 = -x/y ???
and -root3/3 = -x/y??

14. Jan 16, 2008

### rocomath

Your slopes are rationalized while the ones I used are not. It's easier to type into the calculator or to write.

15. Jan 16, 2008

### rocomath

Tell me what your coordinates would be if you were to use those values?

If you plug in those values into your original equation, will it equal 4? Rationalizing isn't always a good thing to do.

16. Jan 16, 2008

### VanKwisH

ahhhhhh ic ........... so root3/3 = 1/root3

17. Jan 16, 2008

### rocomath

Yep, and it's definitely best to use the unrationalized version so that you can also determine what your x-points will at those tangents.

18. Jan 16, 2008

### VanKwisH

y – y1 = m(x – x1) point slope formula right???
so i have the point (-1,root3) from the first slope
now i find the point from the second slope right??
and it should be ( 1,root3) >??

19. Jan 17, 2008

### rocomath

Yes, and since this a circle, only 1 point will change.

20. Jan 17, 2008

### VanKwisH

alright so i plug in those co-ordinates into the point slope equation?? and that should find me my intersect point right?

21. Jan 17, 2008

### VanKwisH

OOO wait crap what about the m

22. Jan 17, 2008

### rocomath

Your not looking for the equation of the intersection. It's just the x-y coordinate.

Set your tangents equal to each other and solve for x. Then find y from either equation.

23. Jan 17, 2008

### VanKwisH

can u show that mathematically cause it doesn't make any sense to me at all

24. Jan 17, 2008

### rocomath

$$y_1^{'}=\frac{-x}{\sqrt3}+\frac{4}{\sqrt3}$$ & $$y_2^{'}=\frac{x}{\sqrt3}+\frac{4}{\sqrt3}$$

$$y_1^{'}=y_2^{'}$$

$$\frac{-x}{\sqrt3}+\frac{4}{\sqrt3}=\frac{x}{\sqrt3}+\frac{4}{\sqrt3}$$

Solve for x.

25. Jan 17, 2008

### VanKwisH

where did the 4 / root 3 come from??