find an equation of tangent line given a point Q(9,5) with center P(6,4)

Question

amistre64 · Answer

tangent to a point is not defined

anonymous · Answer

?

amistre64 · Answer

read your post and see if you can make any sense out of it ....

anonymous · Answer

well i have the standard form which is (x-6)^2+(y-4)^2=square root of 10

amistre64 · Answer

thats something we can work with then :)

anonymous · Answer

thank you

amistre64 · Answer

take its derivative ...

anonymous · Answer

im not in calculus yet

anonymous · Answer

im in pre-calculus

amistre64 · Answer

then we need to find the slope of the line from the Q to P which then makes sense

amistre64 · Answer

the tangent will be the perp slope

anonymous · Answer

okay

anonymous · Answer

show me

anonymous · Answer

center = (6,4)

amistre64 · Answer

Q(9,5) with center P(6,4)
  -9-5                  -9-5
-------               ------
  0,0                     -3,-1 ; slope = y/x:  1/3

anonymous · Answer

point = (9,5)

amistre64 · Answer

in this case; either point doesnt matter; slope between them is still the same

anonymous · Answer

why is -9 and not just 9?

amistre64 · Answer

becasue 9+9 isnt 0

amistre64 · Answer

to get (9,5) to (0,0)  we have to subtract it from itself

amistre64 · Answer

and whatever we do to one point, we do to them all

anonymous · Answer

ugh....

anonymous · Answer

you lost me there

anonymous · Answer

sorry its just that i haven't done math in so long

amistre64 · Answer

thats becasue im doing this my way; how do you find slope between 2 points?

anonymous · Answer

ill medal you each answer

anonymous · Answer

change in y/change in x

amistre64 · Answer

good; then what is our change in y and our change in x from point to point?

anonymous · Answer

so i would do 9-5/5-4 = 4/1 which is equal to just 4

amistre64 · Answer

subtract one point from the other and you end up with the changes

amistre64 · Answer

your slope formula is fine except that you are not placing the points in the right spot

which is why i avoid that method and just step it out....

amistre64 · Answer

Q(9,5)
-P(6,4)
------
    3, 1 ; slope = 1/3

anonymous · Answer

oh sorry

anonymous · Answer

i mean 9-6/5-4 = 3/1 = 3

amistre64 · Answer

your 9-5 should have been 9-6 ... but becasue of the manner in which you are trying to fill in parts of a formula you tend to bring in more confusion along the way

amistre64 · Answer

and your putting your xs over your ys

anonymous · Answer

oh wait 
y/x

anonymous · Answer

so its 5-4/9-6 = 1/3

amistre64 · Answer

subtract your points; then stack y/x ... avoids alot of messups

anonymous · Answer

ugh im so stupid

amistre64 · Answer

1/3 yes is the slope between the points; now flip and negate for a perp

amistre64 · Answer

-3/1 = -3

anonymous · Answer

okay now that i have the slope from the point to the radius what do i do next?

amistre64 · Answer

we want an equation of the tangent line; so we have the slope of the tangent line; which point is the outside of the circle again?

anonymous · Answer

point Q(9,5)
center (6,4)

anonymous · Answer

P(6,4)

amistre64 · Answer

then all we have to do is fill in the exterior point in

y = mx -mPx + Py

y = -3x +3(9) + (5)

amistre64 · Answer

the point slope form of a line; given (Px,Py) and a slope m is

y - Py = m(x-Px)

y  = m(x-Px) + Py

y  = mx -mPx) + Py

amistre64 · Answer

m=-3 in this case and we just fill in our given point (9,5)

anonymous · Answer

y-5=-3(x-9)   ???