anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
so like this?
1 Attachment
anonymous
  • anonymous
YESSSSSSSSS!!!!!!!!!!!!!!
anonymous
  • anonymous
in an xy plane

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amistre64
  • amistre64
since a tangent to the circle is always at 90 degrees to the radius.... this makes a right triangle.
anonymous
  • anonymous
hey how do you draw attachments like that
amistre64
  • amistre64
the hyp is the length of the center of the cirlce to the point and one leg is the radius
amistre64
  • amistre64
i use paint in the windows accessories...save it to my desktop and retrieve it from there
anonymous
  • anonymous
you can draw circles?
amistre64
  • amistre64
1 Attachment
amistre64
  • amistre64
yes; its the oval selection and hold down the shift key while you drag
anonymous
  • anonymous
yes what program is it? windows paint?
anonymous
  • anonymous
ok so we have a point, lets call it (x0,y0) outside a circle
amistre64
  • amistre64
yes; windows generi paint program that comes in every installation of windows since 3.xx
anonymous
  • anonymous
ok i will check that
anonymous
  • anonymous
i want to actually find a general formula here
anonymous
  • anonymous
ok so we have (x0,y0) outside a circle (x-h)^2 + (y-k)^2 = r^2
amistre64
  • amistre64
the general formula is the pythag thrm... to find a vector
amistre64
  • amistre64
find a vector that is parallel to the tangent from the point....
anonymous
  • anonymous
you cant do this analytically
amistre64
  • amistre64
you can ; but then convert it :)
anonymous
  • anonymous
we have y - y0 = m(x-x0) ,
amistre64
  • amistre64
the slope of the tangent to the circle at any given point is f'(x)
anonymous
  • anonymous
also the line from the center of circle to (x,y) is y-k = -1/m (x-h)
anonymous
  • anonymous
the normal i mean
anonymous
  • anonymous
since (x,y) is unknown
amistre64
  • amistre64
given a standard point on the circle for say (Xc,Yc)...
anonymous
  • anonymous
oh then i messed up
amistre64
  • amistre64
the point has to be within a certain domain for an outside point to have a tangent right?
anonymous
  • anonymous
thats possible
anonymous
  • anonymous
there are 3 points given
anonymous
  • 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
  • anonymous
so 4 points
amistre64
  • amistre64
f'(c) that passes thru P(x1,y1) in the field and P1(xc,yc) on the circle
amistre64
  • amistre64
if we view this from the origin...would it be simpler?
amistre64
  • 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
  • anonymous
yes
amistre64
  • amistre64
like this so far
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amistre64
  • amistre64
a' is in the wrong spot lol
anonymous
  • anonymous
so you can do this vectorially? or with derivatives?
amistre64
  • amistre64
you can, but lets get a clear picture first and then add the details :)
amistre64
  • amistre64
the perp lines slope we want is tan(a+b) right?
anonymous
  • anonymous
tan a + b ?
amistre64
  • amistre64
a = cos^-1(CP/r)
amistre64
  • amistre64
yes; from my picture I got a + b = the slope of the line that is perp to the tangent....
amistre64
  • amistre64
and tan(a+b) givs us that slope; so figure the angles... :)
amistre64
  • amistre64
does this makes sense?
anonymous
  • anonymous
no
amistre64
  • amistre64
we center the circle at the origin so we can examine it...
anonymous
  • anonymous
ok
anonymous
  • anonymous
i dont understand the a = arcos (CP/r) business, where does that come from
amistre64
  • 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
  • anonymous
ok , the angle is ?
amistre64
  • amistre64
the angle is cos(a) = P.r/(|P||r|)...dot product of 2 vecotrs
anonymous
  • anonymous
cos a = Ca' / ( CP)
amistre64
  • amistre64
P being the outside point
anonymous
  • anonymous
what is r ?
amistre64
  • amistre64
cos(b) = T.P/(|P||r|)
amistre64
  • amistre64
radius = r.....
anonymous
  • anonymous
cos b =r / TP
amistre64
  • amistre64
to find the tangent..... we use the pythag thrm; |T|=sqrt(|P|^2-r^2)
anonymous
  • anonymous
ok but we want the tangent line equation
anonymous
  • anonymous
|T| is infinity
anonymous
  • anonymous
but that line segment, right comes from pythangrean
anonymous
  • anonymous
and you were doing the inner product, im a bit rusty
amistre64
  • amistre64
back...
anonymous
  • anonymous
hey
amistre64
  • 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
  • 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
  • anonymous
oh i think i have it
anonymous
  • anonymous
if youre given a circle and a point outside a circle
anonymous
  • anonymous
nevermind
amistre64
  • amistre64
...... but i gotta mind lol
amistre64
  • 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
  • anonymous
i guess im confused on a lot of things. where did you get tan (a+b)
anonymous
  • anonymous
where did you get arcos (cP/r)
amistre64
  • amistre64
i was rummaging thru the thoughts in my head.... some were good, others not so good :)
anonymous
  • anonymous
where did point c come from ?
anonymous
  • anonymous
ok youre starting over
amistre64
  • amistre64
c = center; ..... or origin depending on where the circle was
amistre64
  • amistre64
since we dont know the point ot tangency, i figure vectors are a safe way to find it
anonymous
  • anonymous
yeah thats the trouble
amistre64
  • 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
  • amistre64
p is the vector from the origin to the outside point..
amistre64
  • amistre64
we know that if we subtract p from r we get a new vector heading inthe direction of tangency...
amistre64
  • amistre64
it has to match f'(c) in order to determine if 'c' is a match..
anonymous
  • anonymous
one sec
amistre64
  • 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
  • anonymous
ok vector R - P has the same direction as the tangent line , i agree
anonymous
  • anonymous
because P + (R-P) = R , where R is that point (c,f(c)) or the vector to the point
anonymous
  • anonymous
so f ' (c) will , you mean the vector c ? or just
amistre64
  • amistre64
the derivative of the function itself.
anonymous
  • anonymous
so f ' ( < c , f(c) ) >
anonymous
  • anonymous
oh , the derivative of the equation of the circle
amistre64
  • amistre64
the circle is what i had in mind; but the vector might work just as good
anonymous
  • anonymous
2x + 2y * y' = 0
anonymous
  • anonymous
i prefer analytic, my vector skills are shaky :)
anonymous
  • anonymous
so y ' = -x/y
amistre64
  • amistre64
vectors are analytic; just useful when drawn as arrows lol
anonymous
  • anonymous
true
anonymous
  • anonymous
ok i prefer baby algebra ?
amistre64
  • amistre64
lets restrict our selves to half the circle..... for simplicity y = sqrt(r^2 - x^2)
anonymous
  • anonymous
its confusing, because the points , like an n tuple, can be represented as a geometric point, or an arrow from the origin,
anonymous
  • anonymous
ok
anonymous
  • anonymous
so f ' (c) = -c / ( sqrt ( r^2 - c^2 )
anonymous
  • anonymous
you must be good at geometry, im better with algebra/symbols, that sort of thing
anonymous
  • anonymous
so we have x = tv + x0 or something
anonymous
  • anonymous
x = x0 + at , y = y0 + at , but then we will have to remove the t
amistre64
  • 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
  • amistre64
the t is just a scalar..
amistre64
  • amistre64
it can be 1 for all it cares
anonymous
  • anonymous
well i want the final solution to look like y = mx + b
anonymous
  • anonymous
the problem is that (c,f(c) is unknown
amistre64
  • amistre64
c has to meet the criteria set forth.... once it meets it; we know it :)
radar
  • radar
will it meet it today?
amistre64
  • amistre64
lol...depends on how it behaves lol
amistre64
  • amistre64
if anything ; f'(c) is the slope of the line passing thru points (c,f(c)) and (x,y)
anonymous
  • anonymous
ok
amistre64
  • amistre64
y = f'(c)x + B
anonymous
  • anonymous
ok
amistre64
  • 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
  • amistre64
-c -3 --------- = ---- sqrt(4-c^2) c-6
amistre64
  • amistre64
cross multiply... -c(c-6) = -3(sqrt(4-c^2)) ; /-3 -c^2 +6c -------- = sqrt(4-c^2 ; ^2 both sides.. -3
amistre64
  • amistre64
c^4 -12c^3 +36c^2 ------------------ = 4-c^2 9
amistre64
  • amistre64
*9 both sides c^4 -12c^3 +36c^2 = 36 - 9c^2
amistre64
  • amistre64
zero out the rhs... c^4 -12c^3 +9c^2 + 36c^2 -36 = 0 c^4 -12c^3 + 45c^2 -36 = 0
anonymous
  • anonymous
how did you get this -c -3 -3 --------- =----------- sqrt(4-c^2) c-6
anonymous
  • anonymous
there are some algebraic approaches on yahoo answers, but im not sure they are correct
anonymous
  • anonymous
http://answers.yahoo.com/question/index?qid=20090721083008AA1VEba
amistre64
  • amistre64
the derivative at c = -c/sqrt(r^2-c^2) right?
amistre64
  • amistre64
at r = 2....
amistre64
  • amistre64
...... i see a hiccup tho; i used (c,2) instead of (c,f(c)).... gonna have to rewrite it al lol
amistre64
  • amistre64
-c sqrt(4-c^2)-5 ---------- = ------------- sqrt(4-c^2) c-6
amistre64
  • 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
  • amistre64
now ^2 both sides :) 6c^2 -48c +16 = 25(4-c^2) 6c^2 -48c +16 = 100 -25c^2
amistre64
  • amistre64
36c^2....
amistre64
  • amistre64
36c^2 +25c^2 -48c +16 -100 = 0 61c^2 -48c -84 = 0
amistre64
  • amistre64
solve for c
anonymous
  • anonymous
hmmm
anonymous
  • anonymous
well this does seem easier than the yahoo method , its insanely difficult
anonymous
  • anonymous
if you want to read it its here http://answers.yahoo.com/question/index?qid=20090721083008AA1VEba
anonymous
  • anonymous
YOU GET MEDAL for persistence !!!
amistre64
  • amistre64
:) i get from teh wolfram graphing ; that c = -.844 and 1.631
anonymous
  • anonymous
one sec, let me draw this
anonymous
  • anonymous
wait
anonymous
  • anonymous
so you said the direction of R- P where R-P = say is b/a >
anonymous
  • anonymous
i think theres something wrong with that , oh, it only works in 2 dimensions
amistre64
  • 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
  • amistre64
...got some signs wrong lol
anonymous
  • anonymous
so basically you did f ' (c) = slope [ segment RP ]
anonymous
  • anonymous
and we know that f ' (c) from the implicit derivative -x/y
amistre64
  • amistre64
yes
anonymous
  • anonymous
oh well then you didnt really need to use vectors then , i guess
amistre64
  • 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
  • anonymous
in higher dimensions we will need vectors
amistre64
  • amistre64
vector is used to calibrate the slope
anonymous
  • anonymous
but you took the slope of that vector? correct ?
amistre64
  • amistre64
(a^2 +b^2) = A (-2ra) = B [r^2(1-b^2)] = C
amistre64
  • amistre64
correct, the vector x and y form a slope.... so yes
amistre64
  • amistre64
c = thats stuf in the quadraitc formula lol
anonymous
  • anonymous
whats all this Ac^2 +Bc + C (a^2 -b^2) = A (-2ra) = B [r^2(1+b^2)] = C
amistre64
  • amistre64
i took the numbers out that i was using and placed the (a,b) fromthe exteroir point back into it...
amistre64
  • 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
  • anonymous
youre making a general solution
amistre64
  • amistre64
yes :)...its no good if it only used for one point
anonymous
  • anonymous
ok start over, where did you A, B< and C
amistre64
  • 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
  • amistre64
61c^2 -48c -84 = 0 [c] can be solved by teh quadratic formula now since we know a b and c
amistre64
  • amistre64
61 is from a^2 + b^2; 36 +26 = 61
amistre64
  • 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
  • anonymous
ok
amistre64
  • 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
  • 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
  • anonymous
ok starting with a point (a,b)
anonymous
  • 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
  • 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
  • amistre64
if the center is not at the origin; then use transformations to get it there :)
anonymous
  • anonymous
actually this wont work
anonymous
  • anonymous
-c / sqrt ( r^2 -c^2) = ...
anonymous
  • anonymous
the right side could be b - f(c) or f(c) - b
anonymous
  • anonymous
we used (f(c) - b ) / ( c- a)
anonymous
  • anonymous
so you can say, if f(c) > b, then ...
amistre64
  • amistre64
its the same slope regardless of the way its pointing
anonymous
  • anonymous
oh right , duh
anonymous
  • anonymous
so then we get -c / sqrt ( r^2 -c^2) =(f(c) - b ) / ( c- a)
anonymous
  • anonymous
-c / sqrt ( r^2 -c^2) =(srt (r^2 - c^2) - b ) / ( c- a)
amistre64
  • 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
  • anonymous
correct
anonymous
  • anonymous
but i get a simpler derivation
amistre64
  • amistre64
but we have the slope; and we have a point from the vectors...
anonymous
  • 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
  • anonymous
-c (c-a) =(r^2 - c^2) - b sqrt (r^2-c^2)
anonymous
  • anonymous
actually i cross multiplied , ok so far?
amistre64
  • amistre64
good so far
anonymous
  • anonymous
sqrt (r^2 - c^2) = r^2/b - ac/b
amistre64
  • amistre64
i think we broke openstudy... the fonts gotten weird on my end
anonymous
  • anonymous
you can start over, use www.openstudy.com , and click math
anonymous
  • anonymous
r^2 - c ^2 = ( r^2 - ac)^2 / b^2
anonymous
  • 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
  • anonymous
oh i subtracted c^2 from both sides by the way
amistre64
  • amistre64
so did i :)
anonymous
  • anonymous
ok squaring both sides i get r^2*b^2 - b^2 c^2 = r^4 -2r^2ac + a^2 *c^2
anonymous
  • anonymous
now r is given, b and a are given
anonymous
  • anonymous
i didnt get A = a^2 +b ^2 though
anonymous
  • anonymous
still there?

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