how do you find the equation of a tangent line through a point outside a circle.
Suppose the point is (6,-5) and the circle is
(x-1)^2 + (y^2) = 4 , but i want to find a general formula.
we know that two tangents can always be drawn to a circle from any point outside the circle

- anonymous

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- amistre64

so like this?

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- anonymous

YESSSSSSSSS!!!!!!!!!!!!!!

- anonymous

in an xy plane

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## More answers

- amistre64

since a tangent to the circle is always at 90 degrees to the radius.... this makes a right triangle.

- anonymous

hey how do you draw attachments like that

- amistre64

the hyp is the length of the center of the cirlce to the point and one leg is the radius

- amistre64

i use paint in the windows accessories...save it to my desktop and retrieve it from there

- anonymous

you can draw circles?

- amistre64

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- amistre64

yes; its the oval selection and hold down the shift key while you drag

- anonymous

yes
what program is it? windows paint?

- anonymous

ok so we have a point, lets call it (x0,y0) outside a circle

- amistre64

yes; windows generi paint program that comes in every installation of windows since 3.xx

- anonymous

ok i will check that

- anonymous

i want to actually find a general formula here

- anonymous

ok so we have (x0,y0) outside a circle (x-h)^2 + (y-k)^2 = r^2

- amistre64

the general formula is the pythag thrm... to find a vector

- amistre64

find a vector that is parallel to the tangent from the point....

- anonymous

you cant do this analytically

- amistre64

you can ; but then convert it :)

- anonymous

we have y - y0 = m(x-x0) ,

- amistre64

the slope of the tangent to the circle at any given point is f'(x)

- anonymous

also the line from the center of circle to (x,y) is y-k = -1/m (x-h)

- anonymous

the normal i mean

- anonymous

since (x,y) is unknown

- amistre64

given a standard point on the circle for say (Xc,Yc)...

- anonymous

oh then i messed up

- amistre64

the point has to be within a certain domain for an outside point to have a tangent right?

- anonymous

thats possible

- anonymous

there are 3 points given

- anonymous

(h,k) the center of the circle, (xo,yo) point outside circle
(x1,y1) the point on the circle and (x2,y2) the second point on circle that intersects tangent

- anonymous

so 4 points

- amistre64

f'(c) that passes thru P(x1,y1) in the field and P1(xc,yc) on the circle

- amistre64

if we view this from the origin...would it be simpler?

- amistre64

the angle formed by the center to outside point forms an angle [a] with the x axis; the angle between that line and the radius to tangent form [b] right

- anonymous

yes

- amistre64

like this so far

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- amistre64

a' is in the wrong spot lol

- anonymous

so you can do this vectorially? or with derivatives?

- amistre64

you can, but lets get a clear picture first and then add the details :)

- amistre64

the perp lines slope we want is tan(a+b) right?

- anonymous

tan a + b ?

- amistre64

a = cos^-1(CP/r)

- amistre64

yes; from my picture I got a + b = the slope of the line that is perp to the tangent....

- amistre64

and tan(a+b) givs us that slope; so figure the angles... :)

- amistre64

does this makes sense?

- anonymous

no

- amistre64

we center the circle at the origin so we can examine it...

- anonymous

ok

- anonymous

i dont understand the a = arcos (CP/r) business, where does that come from

- amistre64

the angle made by the line from the center to the extPoint makes an angle [a] with the x axis...that angle is cos^-1[]

- anonymous

ok , the angle is ?

- amistre64

the angle is cos(a) = P.r/(|P||r|)...dot product of 2 vecotrs

- anonymous

cos a = Ca' / ( CP)

- amistre64

P being the outside point

- anonymous

what is r ?

- amistre64

cos(b) = T.P/(|P||r|)

- amistre64

radius = r.....

- anonymous

cos b =r / TP

- amistre64

to find the tangent..... we use the pythag thrm;
|T|=sqrt(|P|^2-r^2)

- anonymous

ok but we want the tangent line equation

- anonymous

|T| is infinity

- anonymous

but that line segment, right comes from pythangrean

- anonymous

and you were doing the inner product, im a bit rusty

- amistre64

back...

- anonymous

hey

- amistre64

so if we know tha toutside point ot not; we can assign a vector to it ..p we can also assign a vector to our radius at any given point [c]... r = .
when f'(c) is parallel with r-p..... we have our tangent line

- amistre64

the equation of a tangent line thru any point is given by:
x = x0 + at
y = y0 + bt
z = 0....since we are in the xy plane

- anonymous

oh i think i have it

- anonymous

if youre given a circle and a point outside a circle

- anonymous

nevermind

- amistre64

...... but i gotta mind lol

- amistre64

if we are examine it from the origin; like all good analysts; we move the point and the circle by -x and -y..... make sure to write that in a note so we know were to put it when were done...

- anonymous

i guess im confused on a lot of things. where did you get tan (a+b)

- anonymous

where did you get arcos (cP/r)

- amistre64

i was rummaging thru the thoughts in my head.... some were good, others not so good :)

- anonymous

where did point c come from ?

- anonymous

ok youre starting over

- amistre64

c = center; ..... or origin depending on where the circle was

- amistre64

since we dont know the point ot tangency, i figure vectors are a safe way to find it

- anonymous

yeah thats the trouble

- amistre64

we know the origin; we have vectors splayed out from there. the vector for the radius is simply where c is the point of tangency right?

- amistre64

p is the vector from the origin to the outside point..

- amistre64

we know that if we subtract p from r we get a new vector heading inthe direction of tangency...

- amistre64

it has to match f'(c) in order to determine if 'c' is a match..

- anonymous

one sec

- amistre64

when we find the right [c] we have the f'(c) as the slope for our line; and we have a point to place in it to figure out the b of y=f'(c)x + b

- anonymous

ok vector R - P has the same direction as the tangent line , i agree

- anonymous

because P + (R-P) = R , where R is that point (c,f(c)) or the vector to the point

- anonymous

so f ' (c) will , you mean the vector c ? or just

- amistre64

the derivative of the function itself.

- anonymous

so f ' ( < c , f(c) ) >

- anonymous

oh , the derivative of the equation of the circle

- amistre64

the circle is what i had in mind; but the vector might work just as good

- anonymous

2x + 2y * y' = 0

- anonymous

i prefer analytic, my vector skills are shaky :)

- anonymous

so y ' = -x/y

- amistre64

vectors are analytic; just useful when drawn as arrows lol

- anonymous

true

- anonymous

ok i prefer baby algebra ?

- amistre64

lets restrict our selves to half the circle..... for simplicity
y = sqrt(r^2 - x^2)

- anonymous

its confusing, because the points , like an n tuple, can be represented as a geometric point, or an arrow from the origin,

- anonymous

ok

- anonymous

so f ' (c) = -c / ( sqrt ( r^2 - c^2 )

- anonymous

you must be good at geometry, im better with algebra/symbols, that sort of thing

- anonymous

so we have x = tv + x0 or something

- anonymous

x = x0 + at , y = y0 + at , but then we will have to remove the t

- amistre64

the derivative of u^(1/2) = 1/[2u^(1/2)]
u = 1-x^2 ; du = -2x
the derivative of sqrt(r^2 - x^2)...so i agree with the derivative :)

- amistre64

the t is just a scalar..

- amistre64

it can be 1 for all it cares

- anonymous

well i want the final solution to look like y = mx + b

- anonymous

the problem is that (c,f(c) is unknown

- amistre64

c has to meet the criteria set forth.... once it meets it; we know it :)

- radar

will it meet it today?

- amistre64

lol...depends on how it behaves lol

- amistre64

if anything ; f'(c) is the slope of the line passing thru points (c,f(c)) and (x,y)

- anonymous

ok

- amistre64

y = f'(c)x + B

- anonymous

ok

- amistre64

ive been playing some numbers here... see if this follows...
p =<6,5> ; r = 2
R=
f'(c) = -1/sqrt(2^2 - c^2)
R-p = = (c-6,-3)

- amistre64

-c -3
--------- = ----
sqrt(4-c^2) c-6

- amistre64

cross multiply...
-c(c-6) = -3(sqrt(4-c^2)) ; /-3
-c^2 +6c
-------- = sqrt(4-c^2 ; ^2 both sides..
-3

- amistre64

c^4 -12c^3 +36c^2
------------------ = 4-c^2
9

- amistre64

*9 both sides
c^4 -12c^3 +36c^2 = 36 - 9c^2

- amistre64

zero out the rhs...
c^4 -12c^3 +9c^2 + 36c^2 -36 = 0
c^4 -12c^3 + 45c^2 -36 = 0

- anonymous

how did you get this
-c -3 -3
--------- =-----------
sqrt(4-c^2) c-6

- anonymous

there are some algebraic approaches on yahoo answers, but im not sure they are correct

- anonymous

http://answers.yahoo.com/question/index?qid=20090721083008AA1VEba

- amistre64

the derivative at c = -c/sqrt(r^2-c^2) right?

- amistre64

at r = 2....

- amistre64

...... i see a hiccup tho; i used (c,2) instead of (c,f(c)).... gonna have to rewrite it al lol

- amistre64

-c sqrt(4-c^2)-5
---------- = -------------
sqrt(4-c^2) c-6

- amistre64

-c(c-6) = sqrt(4-c^2)[sqrt(4-c^2)-5]
-c^2 +6c = (4-c^2) -5sqrt(4-c^2)
6c -4
------ = sqrt(4-c^2)
-5

- amistre64

now ^2 both sides :)
6c^2 -48c +16 = 25(4-c^2)
6c^2 -48c +16 = 100 -25c^2

- amistre64

36c^2....

- amistre64

36c^2 +25c^2 -48c +16 -100 = 0
61c^2 -48c -84 = 0

- amistre64

solve for c

- anonymous

hmmm

- anonymous

well this does seem easier than the yahoo method , its insanely difficult

- anonymous

if you want to read it its here
http://answers.yahoo.com/question/index?qid=20090721083008AA1VEba

- anonymous

YOU GET MEDAL for persistence !!!

- amistre64

:) i get from teh wolfram graphing ; that c = -.844 and 1.631

- anonymous

one sec, let me draw this

- anonymous

wait

- anonymous

so you said the direction of R- P where R-P = say is b/a >

- anonymous

i think theres something wrong with that , oh, it only works in 2 dimensions

- amistre64

lets define p as (a,b) thats fine... but we dont want to confuse it for the quadratic....
Ac^2 +Bc + C
(a^2 -b^2) = A
(-2ra) = B
[r^2(1+b^2)] = C

- amistre64

...got some signs wrong lol

- anonymous

so basically you did f ' (c) = slope [ segment RP ]

- anonymous

and we know that f ' (c) from the implicit derivative -x/y

- amistre64

yes

- anonymous

oh well then you didnt really need to use vectors then , i guess

- amistre64

I used vectors to compare it inititally to get a formula for the rest...
(a^2 -b^2) = A
(-2ra) = B
[r^2(1+b^2)] = C

- anonymous

in higher dimensions we will need vectors

- amistre64

vector is used to calibrate the slope

- anonymous

but you took the slope of that vector? correct ?

- amistre64

(a^2 +b^2) = A
(-2ra) = B
[r^2(1-b^2)] = C

- amistre64

correct, the vector x and y form a slope.... so yes

- amistre64

c = thats stuf in the quadraitc formula lol

- anonymous

whats all this
Ac^2 +Bc + C (a^2 -b^2) = A (-2ra) = B [r^2(1+b^2)] = C

- amistre64

i took the numbers out that i was using and placed the (a,b) fromthe exteroir point back into it...

- amistre64

the point of tangency is at c..which is equal to the solution to the quadratic equation formed by ....
(a^2+b^2)c^2 -(2ra)c + [r^2(1-b^2)] = 0

- anonymous

youre making a general solution

- amistre64

yes :)...its no good if it only used for one point

- anonymous

ok start over, where did you A, B< and C

- amistre64

..... when i used the the point (6,5) and a radius of 2... i manipulated the algebra to get a final quadratic form right?

- amistre64

61c^2 -48c -84 = 0
[c] can be solved by teh quadratic formula now since we know a b and c

- amistre64

61 is from a^2 + b^2; 36 +26 = 61

- amistre64

-48 is fron -2ra; -2(2)(6).....gonna have to review tha tone lol
and -84 is from (r^2[1-b^2]): 4(1-25) = 4-100......murmur..

- anonymous

ok

- amistre64

paper is so much easier to work on....
Ac^2 +Bc +C
[a^2+b^2] = A
[-2a(r^2)] = B
[r^2(r^2-b^2)] = C

- amistre64

(a,b)
(6,5) ; r=2
(36+25) = 61 = A
-2(6)(2^2) = -12(4) = -48 = B
4(4-25) = 16 - 100 = -84 = C

- anonymous

ok starting with a point (a,b)

- anonymous

external to the circle, and the point tangent on the circle is (c,f(c))
and the radius is r , center , well we used (0,0)

- amistre64

to find the point [c] of tangency we get:
c = :
ar^2 sqrt[4a^2r^4 - 4(a^2+b^2)(r^2(r^2-b^2))]
------- ( +-) -----------------------------
a^2+b^2 a^2+b^2

- amistre64

if the center is not at the origin; then use transformations to get it there :)

- anonymous

actually this wont work

- anonymous

-c / sqrt ( r^2 -c^2) = ...

- anonymous

the right side could be b - f(c) or f(c) - b

- anonymous

we used (f(c) - b ) / ( c- a)

- anonymous

so you can say, if f(c) > b, then ...

- amistre64

its the same slope regardless of the way its pointing

- anonymous

oh right , duh

- anonymous

so then we get -c / sqrt ( r^2 -c^2) =(f(c) - b ) / ( c- a)

- anonymous

-c / sqrt ( r^2 -c^2) =(srt (r^2 - c^2) - b ) / ( c- a)

- amistre64

the whole point is to get to a point c to find the derivative to use in the equation to the tangent line lol

- anonymous

correct

- anonymous

but i get a simpler derivation

- amistre64

but we have the slope; and we have a point from the vectors...

- anonymous

-c / sqrt ( r^2 -c^2) =(srt (r^2 - c^2) - b ) / ( c- a)
multiply both sides by sqrt ( r^2 - c^2)

- anonymous

-c (c-a) =(r^2 - c^2) - b sqrt (r^2-c^2)

- anonymous

actually i cross multiplied , ok so far?

- amistre64

good so far

- anonymous

sqrt (r^2 - c^2) = r^2/b - ac/b

- amistre64

i think we broke openstudy... the fonts gotten weird on my end

- anonymous

you can start over, use www.openstudy.com , and click math

- anonymous

r^2 - c ^2 = ( r^2 - ac)^2 / b^2

- anonymous

-c (c-a) =(r^2 - c^2) - b sqrt (r^2-c^2)
sqrt (r^2 - c^2) =( r^2 - ac)/b ok so far?

- anonymous

oh i subtracted c^2 from both sides by the way

- amistre64

so did i :)

- anonymous

ok squaring both sides i get
r^2*b^2 - b^2 c^2 = r^4 -2r^2ac + a^2 *c^2

- anonymous

now r is given, b and a are given

- anonymous

i didnt get A = a^2 +b ^2 though

- anonymous

still there?

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