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anonymous
 5 years ago
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
(x1)^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
 5 years ago
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 (x1)^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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0YESSSSSSSSS!!!!!!!!!!!!!!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1since a tangent to the circle is always at 90 degrees to the radius.... this makes a right triangle.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey how do you draw attachments like that

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the hyp is the length of the center of the cirlce to the point and one leg is the radius

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i use paint in the windows accessories...save it to my desktop and retrieve it from there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can draw circles?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1yes; its the oval selection and hold down the shift key while you drag

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes what program is it? windows paint?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so we have a point, lets call it (x0,y0) outside a circle

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1yes; windows generi paint program that comes in every installation of windows since 3.xx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i want to actually find a general formula here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so we have (x0,y0) outside a circle (xh)^2 + (yk)^2 = r^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the general formula is the pythag thrm... to find a vector

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1find a vector that is parallel to the tangent from the point....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you cant do this analytically

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1you can ; but then convert it :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have y  y0 = m(xx0) ,

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the slope of the tangent to the circle at any given point is f'(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also the line from the center of circle to (x,y) is yk = 1/m (xh)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since (x,y) is unknown

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1given a standard point on the circle for say (Xc,Yc)...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the point has to be within a certain domain for an outside point to have a tangent right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there are 3 points given

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1f'(c) that passes thru P(x1,y1) in the field and P1(xc,yc) on the circle

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1if we view this from the origin...would it be simpler?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1a' is in the wrong spot lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you can do this vectorially? or with derivatives?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1you can, but lets get a clear picture first and then add the details :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the perp lines slope we want is tan(a+b) right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1yes; from my picture I got a + b = the slope of the line that is perp to the tangent....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1and tan(a+b) givs us that slope; so figure the angles... :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1does this makes sense?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1we center the circle at the origin so we can examine it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont understand the a = arcos (CP/r) business, where does that come from

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the angle made by the line from the center to the extPoint makes an angle [a] with the x axis...that angle is cos^1[]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the angle is cos(a) = P.r/(Pr)...dot product of 2 vecotrs

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1P being the outside point

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1cos(b) = T.P/(Pr)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1to find the tangent..... we use the pythag thrm; T=sqrt(P^2r^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok but we want the tangent line equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but that line segment, right comes from pythangrean

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and you were doing the inner product, im a bit rusty

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1so if we know tha toutside point ot not; we can assign a vector to it ..p <x,y> we can also assign a vector to our radius at any given point [c]... r = <c,f(c)>. when f'(c) is parallel with rp..... we have our tangent line

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the 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
 5 years ago
Best ResponseYou've already chosen the best response.0if youre given a circle and a point outside a circle

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1...... but i gotta mind lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1if 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
 5 years ago
Best ResponseYou've already chosen the best response.0i guess im confused on a lot of things. where did you get tan (a+b)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where did you get arcos (cP/r)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i was rummaging thru the thoughts in my head.... some were good, others not so good :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where did point c come from ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok youre starting over

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1c = center; ..... or origin depending on where the circle was

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1since we dont know the point ot tangency, i figure vectors are a safe way to find it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah thats the trouble

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1we know the origin; we have vectors splayed out from there. the vector for the radius is simply <c,f(c)> where c is the point of tangency right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1p is the vector from the origin to the outside point..<x,y>

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1we know that if we subtract p from r we get a new vector heading inthe direction of tangency...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1it has to match f'(c) in order to determine if 'c' is a match..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1when 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok vector R  P has the same direction as the tangent line , i agree

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because P + (RP) = R , where R is that point (c,f(c)) or the vector to the point

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so f ' (c) will , you mean the vector c ? or just

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the derivative of the function itself.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so f ' ( < c , f(c) ) >

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh , the derivative of the equation of the circle

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the circle is what i had in mind; but the vector might work just as good

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i prefer analytic, my vector skills are shaky :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1vectors are analytic; just useful when drawn as arrows lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok i prefer baby algebra ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1lets restrict our selves to half the circle..... for simplicity y = sqrt(r^2  x^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its confusing, because the points , like an n tuple, can be represented as a geometric point, or an arrow from the origin,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so f ' (c) = c / ( sqrt ( r^2  c^2 )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you must be good at geometry, im better with algebra/symbols, that sort of thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we have x = tv + x0 or something

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x = x0 + at , y = y0 + at , but then we will have to remove the t

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the derivative of u^(1/2) = 1/[2u^(1/2)] u = 1x^2 ; du = 2x the derivative of sqrt(r^2  x^2)...so i agree with the derivative :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the t is just a scalar..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1it can be 1 for all it cares

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well i want the final solution to look like y = mx + b

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the problem is that (c,f(c) is unknown

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1c has to meet the criteria set forth.... once it meets it; we know it :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1lol...depends on how it behaves lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1if anything ; f'(c) is the slope of the line passing thru points (c,f(c)) and (x,y)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1ive been playing some numbers here... see if this follows... p =<6,5> ; r = 2 R=<c,f(c)> f'(c) = 1/sqrt(2^2  c^2) Rp = <c6,25> = (c6,3)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1c 3  =  sqrt(4c^2) c6

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1cross multiply... c(c6) = 3(sqrt(4c^2)) ; /3 c^2 +6c  = sqrt(4c^2 ; ^2 both sides.. 3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1c^4 12c^3 +36c^2  = 4c^2 9

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1*9 both sides c^4 12c^3 +36c^2 = 36  9c^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1zero out the rhs... c^4 12c^3 +9c^2 + 36c^2 36 = 0 c^4 12c^3 + 45c^2 36 = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get this c 3 3  = sqrt(4c^2) c6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there are some algebraic approaches on yahoo answers, but im not sure they are correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://answers.yahoo.com/question/index?qid=20090721083008AA1VEba

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the derivative at c = c/sqrt(r^2c^2) right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1...... i see a hiccup tho; i used (c,2) instead of (c,f(c)).... gonna have to rewrite it al lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1c sqrt(4c^2)5  =  sqrt(4c^2) c6

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1c(c6) = sqrt(4c^2)[sqrt(4c^2)5] c^2 +6c = (4c^2) 5sqrt(4c^2) 6c 4  = sqrt(4c^2) 5

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1now ^2 both sides :) 6c^2 48c +16 = 25(4c^2) 6c^2 48c +16 = 100 25c^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.136c^2 +25c^2 48c +16 100 = 0 61c^2 48c 84 = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well this does seem easier than the yahoo method , its insanely difficult

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you want to read it its here http://answers.yahoo.com/question/index?qid=20090721083008AA1VEba

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0YOU GET MEDAL for persistence !!!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1:) i get from teh wolfram graphing ; that c = .844 and 1.631

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one sec, let me draw this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you said the direction of R P where RP = <a,b> say is b/a >

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think theres something wrong with that , oh, it only works in 2 dimensions

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1lets 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
 5 years ago
Best ResponseYou've already chosen the best response.1...got some signs wrong lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so basically you did f ' (c) = slope [ segment RP ]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and we know that f ' (c) from the implicit derivative x/y

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh well then you didnt really need to use vectors then , i guess

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1I 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
 5 years ago
Best ResponseYou've already chosen the best response.0in higher dimensions we will need vectors

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1vector <ca, f(c)b> is used to calibrate the slope

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but you took the slope of that vector? correct ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1(a^2 +b^2) = A (2ra) = B [r^2(1b^2)] = C

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1correct, the vector x and y form a slope.... so yes

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1c = thats stuf in the quadraitc formula lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whats all this Ac^2 +Bc + C (a^2 b^2) = A (2ra) = B [r^2(1+b^2)] = C

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i took the numbers out that i was using and placed the (a,b) fromthe exteroir point back into it...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the 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(1b^2)] = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0youre making a general solution

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1yes :)...its no good if it only used for one point

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok start over, where did you A, B< and C

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1..... 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
 5 years ago
Best ResponseYou've already chosen the best response.161c^2 48c 84 = 0 [c] can be solved by teh quadratic formula now since we know a b and c

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.161 is from a^2 + b^2; 36 +26 = 61

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.148 is fron 2ra; 2(2)(6).....gonna have to review tha tone lol and 84 is from (r^2[1b^2]): 4(125) = 4100......murmur..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1paper is so much easier to work on.... Ac^2 +Bc +C [a^2+b^2] = A [2a(r^2)] = B [r^2(r^2b^2)] = C

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1(a,b) (6,5) ; r=2 (36+25) = 61 = A 2(6)(2^2) = 12(4) = 48 = B 4(425) = 16  100 = 84 = C

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok starting with a point (a,b)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0external 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
 5 years ago
Best ResponseYou've already chosen the best response.1to find the point [c] of tangency we get: c = : ar^2 sqrt[4a^2r^4  4(a^2+b^2)(r^2(r^2b^2))]  ( +)  a^2+b^2 a^2+b^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1if the center is not at the origin; then use transformations to get it there :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually this wont work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0c / sqrt ( r^2 c^2) = ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the right side could be b  f(c) or f(c)  b

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we used (f(c)  b ) / ( c a)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you can say, if f(c) > b, then ...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1its the same slope regardless of the way its pointing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so then we get c / sqrt ( r^2 c^2) =(f(c)  b ) / ( c a)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0c / sqrt ( r^2 c^2) =(srt (r^2  c^2)  b ) / ( c a)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the whole point is to get to a point c to find the derivative to use in the equation to the tangent line lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i get a simpler derivation

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1but we have the slope; and we have a point from the vectors...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0c / sqrt ( r^2 c^2) =(srt (r^2  c^2)  b ) / ( c a) multiply both sides by sqrt ( r^2  c^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0c (ca) =(r^2  c^2)  b sqrt (r^2c^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually i cross multiplied , ok so far?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt (r^2  c^2) = r^2/b  ac/b

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1i think we broke openstudy... the fonts gotten weird on my end

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can start over, use www.openstudy.com , and click math

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0r^2  c ^2 = ( r^2  ac)^2 / b^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0c (ca) =(r^2  c^2)  b sqrt (r^2c^2) sqrt (r^2  c^2) =( r^2  ac)/b ok so far?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i subtracted c^2 from both sides by the way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok squaring both sides i get r^2*b^2  b^2 c^2 = r^4 2r^2ac + a^2 *c^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now r is given, b and a are given

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i didnt get A = a^2 +b ^2 though
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