Question

Write the equation of the line that is tangent to the circle (x + 6)2 + (y + 4)2 = 25 at the point (-9, -8).

Answer 1

I really would love an explanation, I check google and people get numbers that seem to come out of no where

Answer 2

There is a neat trick in solving these, write the equation like this

\[(x_1+6)(x+6)+(y_1+4)(y+4)=25 \]
Now match the x_1, y_1 value with the coordinates of your point and evaluate

Answer 3

that's a great trick @Spacelimbus
never seen it before

Answer 4

okay, so I get (-3)(x+6) + (-4)(x+4) = 25
\so do I distribute ?

Answer 5

http://www.wolframalpha.com/input/?i=graph+%28x%2B6%29%5E2%2B%28y%2B4%29%5E2%3D25+and+4y%2B3x%3D-59

Answer 6

exactly @Whereisalexis2u

Answer 7

opps theres a y next to 4

Answer 8

@TuringTest, it's really cool, I lost the proof long ago though heehe

Answer 9

all my answer options have 3/4x or 4/3 x

Answer 10

Hmm @Whereisalexis2u, did I read your equation right though? or did you correct something in the meanwhile? You can check my wolfram graphing link above, it's obviously correct, my solution is:

\[ 4y+3x=-59\] so it doesn't go through the origin, but it would be -3/4x

Answer 11

from there I get 59/3 or 4 ones negative and postive.
Youre right however, all the numbers you put down are there

Answer 12

just in a different way.

Answer 13

hmm strange *smiles* but okay then (-:

Answer 14

there should only be one tangent though, not two, two would make no sense.

Answer 15

it's after all a unique point on the circle, therefore the tangent will be unique.

Answer 16

no im saying my answer option is between 59/3 or 59/4
59/3 is postive and 59/4 is negative

Answer 17

oh ok
\[ y = -\frac{3}{4}x - \frac{59}{4} \]

Answer 18

I dont get why its that though.

Answer 19

this whole sections has been a blur to me

Answer 20

some are tricks, a lot is geometry, often it really helps if you try to make a picture of it, it's annoying but it helps.

Answer 21

another way would be implicit differentiation and then derive the slope from that.