Find the equation for the family of lines tangent to the circle with center at the origin and radius 3.

- Shayaan_Mustafa

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- katieb

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- Shayaan_Mustafa

I know, use the equation y=mx+c. But how?

- Hero

First draw the circle

- Shayaan_Mustafa

|dw:1332964463602:dw|

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## More answers

- Hero

Yes, but, well
I also believe we should apply equation of a circle as well to this

- Shayaan_Mustafa

tangent is perpendicular to the radius of the circle. and slope of perpendicular line are m1=-1/m2. right??

- Hero

yes

- Shayaan_Mustafa

let us simplify the question first.
we just find the equation of tangent line and then find the family. OK

- Hero

Let me get back to you regarding this question. I think they want more than just the equation of a tangent line at one point on the circle. I think they want an equation of a tangent for any point on the circle

- Shayaan_Mustafa

yes yes. exactly. family of line implies this idea as you said.
but if we find equation for one tangent the we can find equation for any tangent. family too.

- Shayaan_Mustafa

did you get my idea?

- Hero

if you found an equation for one tangent, you would not be able to find it for another tangent. That would only work if (3,0), (-3,0)(0,3)(0,-3) were the points

- Hero

You would already have to know the points in advance for any other point on the circle in order to find the equations of tangent lines

- Shayaan_Mustafa

look Hero.
y=-x+c is the family of line which has same slope.
do you know this idea?
If so then why not for above question?

- Hero

What concepts are you studying in class regarding this? Are you working from a textbook? If so, tell me the name of the section you're working on.

- Shayaan_Mustafa

Calculus by howard anton. 7th edition

- Hero

Oh, I see. This is calculus. I thought you were doing high school geometry

- Shayaan_Mustafa

it is university level.

- TuringTest

our circle is\[x^2+y^2=9\]the formula for a tangent line at a point 'a' is\[y-y(a)=y'(a)(x-a)\]I think we should differentiate implicitely to get y'

- Shayaan_Mustafa

no. differentiation is not include up to this section. so we can't differentiate it.

- TuringTest

eh? um....
then how is it calculus?

- Shayaan_Mustafa

look look..
it is 1st chapter named functions
and differentiation is starting from chap#3

- Shayaan_Mustafa

no one can help me on this problem :(

- TuringTest

ok um...
let's see if we can be creative based on what we know from functions and trig
the following are tangent lines to the circle\[x=\pm3\]\[y=\pm3\]that covers four sides|dw:1332966190330:dw|now we need to develop some formulas for the areas in between the top, bottom, left, and right

- TuringTest

hm...|dw:1332966361060:dw|here is a tangent line at some mystery point, I wonder if we can divine a formula for the slope of it

- Shayaan_Mustafa

heheheheh... divine..

- TuringTest

|dw:1332966453442:dw|maybe we can use some trig here...

- Hero

Good luck with that

- TuringTest

yeah, maybe a different line of reasoning
let me start a little earlier

- Shayaan_Mustafa

OK OK guys. I got my answer. I have solved the mystry. Come here and see me.

- TuringTest

|dw:1332966712135:dw|the slope of the black line is...
gotta go anyway

- Hero

Anybody wanna go on vyew?

- Shayaan_Mustafa

The circle is given by\[x _{o}^{2}+y _{o}^{2}=9\]right?

- Hero

okay....

- Shayaan_Mustafa

as slope of the radius is given by \[y _{o}/x _{o}\]in general
right?

- Shayaan_Mustafa

so the slope of the tangent line which is also perpendicular to the radius of the circle has the slope of \[-x _{o}/y _{o}\], right?

- Hero

This is all general knowledge stuff Shayaan. The problem I had was I didn't exactly know the specific approach to get to the solution. But I know all of these elements.

- Shayaan_Mustafa

equation for the tangent line is given as\[y=mx+c\]put slope of tangent line we get,\[y=(-x _{o}/y _{o})x+c\]

- Shayaan_Mustafa

\[y _{o}=\pm \sqrt{9-x _{o}^{2}}\]

- Shayaan_Mustafa

now I am confused.

- Shayaan_Mustafa

sorry guys now i am unable to solve this stuff.

- Hero

Just continue simplifying

- Hero

difference of squares

- Hero

assuming everything you did is right. Or maybe just leave the answer as is

- Shayaan_Mustafa

let me solve it again then I post it. thnx for contribution.

- anonymous

is this the question you whant me to answer?

- Shayaan_Mustafa

yes myko.

- anonymous

k

- anonymous

Use the parametric equation for the circle
C = (3cost,3sent) where 0<=t<=2Pi
the equation for the tangent unit vector to the circle would be
t = (-3sent,3cost)/3sqrt2
so the tangent at the point (a,b) would be (x,y) = (a,b)+pt where p goes from 0 to infinity

- Shayaan_Mustafa

thnx.

- anonymous

yw

- TuringTest

Sorry I had something to attend to....
I would perhaps have said something like\[y-y_0=\frac{y_0}{x_0}(x-x_0)=\frac{y_0}{x_0}x-y_0\implies y=\frac{y_0}{x_0}x\]and we have that\[y_0=\pm\sqrt{9-x_0^2}\]so sub that in and we get\[y=\pm\sqrt{(\frac{3}{x_0})^2-1}\]if you want to avoid parametric nonsense
not sure if they want you top break it into a step function though, because the \(\pm\) makes it not a function. oh well..

- Shayaan_Mustafa

hmmm.. this seems familiar much.

- TuringTest

oh it just says "find the equation" so I would leave it with the \(\pm\)

- Shayaan_Mustafa

yes..

- TuringTest

because it does not require the family of lines to be described by a function

- TuringTest

oh I forgot the x

- Shayaan_Mustafa

Thnx a lot Turning Test and everyone who helped me on this complex question.

- TuringTest

\[y=\pm x\sqrt{(\frac{3}{x_0})^2-1}\]

- TuringTest

you're welcome :)

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