Question

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

Answer 1

so like this?

Answer 2

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

Answer 3

in an xy plane

Answer 4

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

Answer 5

hey how do you draw attachments like that

Answer 6

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

Answer 7

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

Answer 8

you can draw circles?

Answer 9

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

Answer 10

yes
what program is it? windows paint?

Answer 11

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

Answer 12

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

Answer 13

ok i will check that

Answer 14

i want to actually find a general formula here

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

you cant do this analytically

Answer 19

you can ; but then convert it :)

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

the normal i mean

Answer 24

since (x,y) is unknown

Answer 25

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

Answer 26

oh then i messed up

Answer 27

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

Answer 28

thats possible

Answer 29

there are 3 points given

Answer 30

(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

Answer 31

so 4 points

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

yes

Answer 36

like this so far

Answer 37

a' is in the wrong spot lol

Answer 38

so you can do this vectorially? or with derivatives?

Answer 39

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

Answer 40

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

Answer 41

tan a + b ?

Answer 42

a = cos^-1(CP/r)

Answer 43

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

Answer 44

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

Answer 45

does this makes sense?

Answer 46

no

Answer 47

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

Answer 48

ok

Answer 49

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

Answer 50

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[]

Answer 51

ok , the angle is ?

Answer 52

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

Answer 53

cos a = Ca' / ( CP)

Answer 54

P being the outside point

Answer 55

what is r ?

Answer 56

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

Answer 57

radius = r.....

Answer 58

cos b =r / TP

Answer 59

to find the tangent..... we use the pythag thrm;

|T|=sqrt(|P|^2-r^2)

Answer 60

ok but we want the tangent line equation

Answer 61

|T| is infinity

Answer 62

but that line segment, right comes from pythangrean

Answer 63

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

Answer 64

back...

Answer 65

hey

Answer 66

so 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 r-p..... we have our tangent line

Answer 67

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

Answer 68

oh i think i have it

Answer 69

if youre given a circle and a point outside a circle

Answer 70

nevermind

Answer 71

...... but i gotta mind lol

Answer 72

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...

Answer 73

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

Answer 74

where did you get arcos (cP/r)

Answer 75

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

Answer 76

where did point c come from ?

Answer 77

ok youre starting over

Answer 78

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

Answer 79

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

Answer 80

yeah thats the trouble

Answer 81

we 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?

Answer 82

p is the vector from the origin to the outside point..<x,y>

Answer 83

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

Answer 84

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

Answer 85

one sec

Answer 86

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

Answer 87

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

Answer 88

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

Answer 89

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

Answer 90

the derivative of the function itself.

Answer 91

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

Answer 92

oh , the derivative of the equation of the circle

Answer 93

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

Answer 94

2x + 2y * y' = 0

Answer 95

i prefer analytic, my vector skills are shaky :)

Answer 96

so y ' = -x/y