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in an xy plane
since a tangent to the circle is always at 90 degrees to the radius.... this makes a right triangle.
hey how do you draw attachments like that
the hyp is the length of the center of the cirlce to the point and one leg is the radius
i use paint in the windows accessories...save it to my desktop and retrieve it from there
you can draw circles?
yes; its the oval selection and hold down the shift key while you drag
yes what program is it? windows paint?
ok so we have a point, lets call it (x0,y0) outside a circle
yes; windows generi paint program that comes in every installation of windows since 3.xx
ok i will check that
i want to actually find a general formula here
ok so we have (x0,y0) outside a circle (x-h)^2 + (y-k)^2 = r^2
the general formula is the pythag thrm... to find a vector
find a vector that is parallel to the tangent from the point....
you cant do this analytically
you can ; but then convert it :)
we have y - y0 = m(x-x0) ,
the slope of the tangent to the circle at any given point is f'(x)
also the line from the center of circle to (x,y) is y-k = -1/m (x-h)
the normal i mean
since (x,y) is unknown
given a standard point on the circle for say (Xc,Yc)...
oh then i messed up
the point has to be within a certain domain for an outside point to have a tangent right?
there are 3 points given
(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
so 4 points
f'(c) that passes thru P(x1,y1) in the field and P1(xc,yc) on the circle
if we view this from the origin...would it be simpler?
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
a' is in the wrong spot lol
so you can do this vectorially? or with derivatives?
you can, but lets get a clear picture first and then add the details :)
the perp lines slope we want is tan(a+b) right?
tan a + b ?
a = cos^-1(CP/r)
yes; from my picture I got a + b = the slope of the line that is perp to the tangent....
and tan(a+b) givs us that slope; so figure the angles... :)
does this makes sense?
we center the circle at the origin so we can examine it...
i dont understand the a = arcos (CP/r) business, where does that come from
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
ok , the angle is ?
the angle is cos(a) = P.r/(|P||r|)...dot product of 2 vecotrs
cos a = Ca' / ( CP)
P being the outside point
what is r ?
cos(b) = T.P/(|P||r|)
radius = r.....
cos b =r / TP
to find the tangent..... we use the pythag thrm; |T|=sqrt(|P|^2-r^2)
ok but we want the tangent line equation
|T| is infinity
but that line segment, right comes from pythangrean
and you were doing the inner product, im a bit rusty
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
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
oh i think i have it
if youre given a circle and a point outside a circle
...... but i gotta mind lol
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...
i guess im confused on a lot of things. where did you get tan (a+b)
where did you get arcos (cP/r)
i was rummaging thru the thoughts in my head.... some were good, others not so good :)
where did point c come from ?
ok youre starting over
c = center; ..... or origin depending on where the circle was
since we dont know the point ot tangency, i figure vectors are a safe way to find it
yeah thats the trouble
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?
p is the vector from the origin to the outside point..
we know that if we subtract p from r we get a new vector heading inthe direction of tangency...
it has to match f'(c) in order to determine if 'c' is a match..
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
ok vector R - P has the same direction as the tangent line , i agree
because P + (R-P) = R , where R is that point (c,f(c)) or the vector to the point
so f ' (c) will , you mean the vector c ? or just
the derivative of the function itself.
so f ' ( < c , f(c) ) >
oh , the derivative of the equation of the circle
the circle is what i had in mind; but the vector might work just as good
2x + 2y * y' = 0
i prefer analytic, my vector skills are shaky :)
so y ' = -x/y
vectors are analytic; just useful when drawn as arrows lol
ok i prefer baby algebra ?
lets restrict our selves to half the circle..... for simplicity y = sqrt(r^2 - x^2)
its confusing, because the points , like an n tuple, can be represented as a geometric point, or an arrow from the origin,
so f ' (c) = -c / ( sqrt ( r^2 - c^2 )
you must be good at geometry, im better with algebra/symbols, that sort of thing
so we have x = tv + x0 or something
x = x0 + at , y = y0 + at , but then we will have to remove the t
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 :)
the t is just a scalar..
it can be 1 for all it cares
well i want the final solution to look like y = mx + b
the problem is that (c,f(c) is unknown
c has to meet the criteria set forth.... once it meets it; we know it :)
will it meet it today?
lol...depends on how it behaves lol
if anything ; f'(c) is the slope of the line passing thru points (c,f(c)) and (x,y)
y = f'(c)x + B
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)
-c -3 --------- = ---- sqrt(4-c^2) c-6
cross multiply... -c(c-6) = -3(sqrt(4-c^2)) ; /-3 -c^2 +6c -------- = sqrt(4-c^2 ; ^2 both sides.. -3
c^4 -12c^3 +36c^2 ------------------ = 4-c^2 9
*9 both sides c^4 -12c^3 +36c^2 = 36 - 9c^2
zero out the rhs... c^4 -12c^3 +9c^2 + 36c^2 -36 = 0 c^4 -12c^3 + 45c^2 -36 = 0
how did you get this -c -3 -3 --------- =----------- sqrt(4-c^2) c-6
there are some algebraic approaches on yahoo answers, but im not sure they are correct
the derivative at c = -c/sqrt(r^2-c^2) right?
at r = 2....
...... i see a hiccup tho; i used (c,2) instead of (c,f(c)).... gonna have to rewrite it al lol
-c sqrt(4-c^2)-5 ---------- = ------------- sqrt(4-c^2) c-6
-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
now ^2 both sides :) 6c^2 -48c +16 = 25(4-c^2) 6c^2 -48c +16 = 100 -25c^2
36c^2 +25c^2 -48c +16 -100 = 0 61c^2 -48c -84 = 0
solve for c
well this does seem easier than the yahoo method , its insanely difficult
if you want to read it its here http://answers.yahoo.com/question/index?qid=20090721083008AA1VEba
YOU GET MEDAL for persistence !!!
:) i get from teh wolfram graphing ; that c = -.844 and 1.631
one sec, let me draw this
i think theres something wrong with that , oh, it only works in 2 dimensions
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
...got some signs wrong lol
so basically you did f ' (c) = slope [ segment RP ]
and we know that f ' (c) from the implicit derivative -x/y
oh well then you didnt really need to use vectors then , i guess
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
in higher dimensions we will need vectors
is used to calibrate the slope
but you took the slope of that vector? correct ?
(a^2 +b^2) = A (-2ra) = B [r^2(1-b^2)] = C
correct, the vector x and y form a slope.... so yes
c = thats stuf in the quadraitc formula lol
whats all this Ac^2 +Bc + C (a^2 -b^2) = A (-2ra) = B [r^2(1+b^2)] = C
i took the numbers out that i was using and placed the (a,b) fromthe exteroir point back into it...
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
youre making a general solution
yes :)...its no good if it only used for one point
ok start over, where did you A, B< and C
..... when i used the the point (6,5) and a radius of 2... i manipulated the algebra to get a final quadratic form right?
61c^2 -48c -84 = 0 [c] can be solved by teh quadratic formula now since we know a b and c
61 is from a^2 + b^2; 36 +26 = 61
-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..
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
(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
ok starting with a point (a,b)
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)
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
if the center is not at the origin; then use transformations to get it there :)
actually this wont work
-c / sqrt ( r^2 -c^2) = ...
the right side could be b - f(c) or f(c) - b
we used (f(c) - b ) / ( c- a)
so you can say, if f(c) > b, then ...
its the same slope regardless of the way its pointing
oh right , duh
so then we get -c / sqrt ( r^2 -c^2) =(f(c) - b ) / ( c- a)
-c / sqrt ( r^2 -c^2) =(srt (r^2 - c^2) - b ) / ( c- a)
the whole point is to get to a point c to find the derivative to use in the equation to the tangent line lol
but i get a simpler derivation
but we have the slope; and we have a point from the vectors...
-c / sqrt ( r^2 -c^2) =(srt (r^2 - c^2) - b ) / ( c- a) multiply both sides by sqrt ( r^2 - c^2)
-c (c-a) =(r^2 - c^2) - b sqrt (r^2-c^2)
actually i cross multiplied , ok so far?
good so far
sqrt (r^2 - c^2) = r^2/b - ac/b
i think we broke openstudy... the fonts gotten weird on my end
you can start over, use www.openstudy.com , and click math
r^2 - c ^2 = ( r^2 - ac)^2 / b^2
-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?
oh i subtracted c^2 from both sides by the way
so did i :)
ok squaring both sides i get r^2*b^2 - b^2 c^2 = r^4 -2r^2ac + a^2 *c^2
now r is given, b and a are given
i didnt get A = a^2 +b ^2 though