Finding intersections of tangents on circles

[VanKwisH]

Homework Statement

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

Homework Equations

The Attempt at a Solution

how do i even do this?

 

[rocomath]

What is the equation of your circle?

 

[VanKwisH]

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

 

[rocomath]

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

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.

 

[VanKwisH]

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

 

[rocomath]

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

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

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

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

 

[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

 

[rocomath]

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

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

"Implicit differentiation"

 

[VanKwisH]

AHHHHHHHHHH alrighty 1 sec let me calculate this

 

[VanKwisH]

okay

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

 

[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.

 

[VanKwisH]

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

 

[VanKwisH]

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

 

[rocomath]

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

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

 

[rocomath]

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

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.

 

[VanKwisH]

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

 

[rocomath]

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

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.

 

[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) >??

 

[rocomath]

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) >??

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

 

[VanKwisH]

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

 

[VanKwisH]

OOO wait crap what about the m

 

[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.

 

[VanKwisH]

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

 

[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.

 

[VanKwisH]

where did the 4 / root 3 come from??

 

[rocomath]

$$y-y_1=y'(x-x_1)$$

For coordinate $$(1,\sqrt3)$$

$$y-\sqrt3=\frac{-1}{\sqrt3}(x-1)$$

$$y=\frac{-x}{\sqrt3}+\frac{1}{\sqrt3}+\sqrt3$$

Get a common denominator for your constants and add.

 

[VanKwisH]

oooooooo alrighty damnnn how would i solve for x ? cross multiply and isolate?

 

[rocomath]

oooooooo alrighty damnnn how would i solve for x ? cross multiply and isolate?

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

The $$\frac{4}{\sqrt3}$$ terms cancel out, so you're left with $$\frac{-x}{\sqrt3}=\frac{x}{\sqrt3}$$ now set it equal to 0 and solve for x. Or if you notice, that there is only 1 value that makes this true ...

 

[VanKwisH]

apparently it's at o .... but isn't that impossibile?

 

[VanKwisH]

or am i doing something wrong

 

[rocomath]

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.

 

[VanKwisH]

wtf i am completely lost now ... sorry for being a retard but how is the approx value 2.31

 

[VanKwisH]

mann so many different things and i can't seem to see wtf is going on

 

[rocomath]

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

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

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

$$y_1^{'}=y_2^{'}=\frac{4}{\sqrt3}\approx 2.31$$

 

[VanKwisH]

ooo ic u got it from 4/root 3 ... but i thought we canceled it so it was only
-x/root3 = x/root3