Question

find the equation of a line with a slope of -1 which is tangent to the circle x^2 + y^2 = 9

Answer 1

yohcandoit.

Answer 2

Find the radius of the circle first, then the point where the line hits the circle. then you'll have the point and a slope, then you can find the equation of the line.

Answer 3

the radius would be 3. but how would you find where the line hits the circle?

Answer 4

The radius is the distance. You can use the distance formula to help you find it

Answer 5

You already have the center point, right?

Answer 6

the distance from the origin :O how would that help? the center point must be (0,0)

Answer 7

In your distance formula you have d = 3, x = 0, y = 0, x_1 = 3, the only thing left to find is y_2

Answer 8

huh? I'm not sure I remember the distance formula.

Answer 9

\[d = \sqrt{(x_2 - x_1)^2+(y_2-y_1)^2}\]

Answer 10

but then y = 0.. I don't see how this helps/

Answer 11

y doesn't equal zero

Answer 12

I ended up with 3 = root (0-3)^2 +(y-0)^
so -3+y = 3

Answer 13

y = 6

Answer 14

oh yeah huh...

Answer 15

so x^2 + 6^2 = 9..
how am I closer to finding the line?

Answer 16

hoh crap. I got a gym appointment in 15. thanks for your help, but I still don't get it.

Answer 17

Okay

Answer 18

Hey mybrainvsme please check you question again and see if they give a point where its tangent to on the circle

Answer 19

No nayeaddo, you have to find the point

Answer 20

Well you have to find the equation of the line and all you need is the y-intercept because you already have the slope. I'm not sure if I understand your method of solving for that point

Answer 21

I'll show you

Answer 22

okay

Answer 23

\[x^2 + y^2= \sqrt{(x_2 - x_1)^2+(y_2-y_1)^2}\]

Kinda like systems of equations we have to set the circle equal to the distance formula. That how you start it off. Or rather that is one approach

Answer 24

we can replace the x^2 + y^2 with 9 and x_1, y_1 with 0

Answer 25

The point is 3,0

Answer 26

I found it without even doing any calculations

Answer 27

(3,0) ?

Answer 28

Yes

Answer 29

So its tangent to the point (3,0) ?

Answer 30

as far as I know

Answer 31

That's the only point that seems to work

Answer 32

yes but that doesn't really hold true then the line has a undefined slope
(3,0) is an x-intercept so if the line should be tangent to that point on the circle then its a straight line

Answer 33

I'm thinking that maybe calculus is involved here by taking implicit differentiation of function...I wanted to ask the questioner what class this question came from...did you get what I said before?

Answer 34

Yeah, I'm just an amateur, but I don't think I was totally off

Answer 35

The point is definitely not 3,0 I realize, however

Answer 36

yeah...the question itself seems to be a bit vague most of the tangent line problems you're giving a specific point where line touches function..hmmm

Answer 37

Okay, I have it

Answer 38

It's easy, but it's not easy to explain on here.

Answer 39

Here's my solution because a circle is not a function and in our case its symmetrical across both the x and y axis so I split them up into two equations:
\[y = \sqrt{9 - x ^{2}} ; y = -\sqrt{9 -x ^{2}}\]

Answer 40

Okay, you first

Answer 41

You take the first derivative of the first function and set it equal to the slope of the slope of the line which is -1.

Answer 42

Yeah, that's good

Answer 43

There's another way to solve it without using calculus

Answer 44

The tangents form triangles with the with the x-axis with short sides 3 so hypotenuse is sqrt(18) giving plus or minus 3 sqrt 2 So lines are y = -x +- 3sqrt2 http://www.wolframalpha.com/input/?i=plot+x^2+%2B+y^2+%3D+9%2Cy%3D-x-3sqrt2%2Cy%3D-x%2B3sqrt2+%2Cx%3D0%2Cy%3D0%2Cy%3Dx%2C+x%3D-5..5%2Cy%3D-5..5

Answer 45

estudier has it

Answer 46

you get\[-x / \sqrt{9 - x ^{2}} = -1\]

when you solve for x you get sqrt(4.5) and y is also sqrt(4.5) okay whats your way

Answer 47

estudier already posted it

Answer 48

But that's not even the way I did it

Answer 49

Geometry is king....

Answer 50

yeah but calculus makes it funner lol I got the same answer 3sqrt(2) = 2sqrt(4.5) for the intercept

Answer 51

The slope is negative one, the radius is three. the line from the center of the circle to the point is perpendicular to the slope of the line

Answer 52

I don't I like using calculus

Answer 53

so the equation perpendicular to the equation of the line is y = x

Answer 54

substitute x into the equation of the circle, replacing y, then solve for x

Answer 55

you get x = 3sqrt2/2

Answer 56

And then you can continue solving from there

Answer 57

yeah good approach too because y = x is a one-to-one function so its convenient to plug in x into equation of circle

Answer 58

:o), I'm glad you understood what I was attempting to explain

Answer 59

yes, have you studied calculus?

Answer 60

Yes, I have, but I'm rusty

Answer 61

oh okay

Answer 62

As u can see, not required...

Answer 63

Precisely

Answer 64

If I can solve it without using calculus, I do

Answer 65

Or at least try to

Answer 66

yeah, the only reason I brought it up is I wanted to find out which class the questioner was taking or from what context he posed the question because they teach tangent lines in calculus I so my approach would have been correct in that case...but I also understand what you guys are saying

Answer 67

yeah..sometimes you forget the basics

Answer 68

they are supposed to teach tangent lines in geometry...

Answer 69

They're supposed to teach a lot of things in geometry, but they skip a lot of things

Answer 70

My high school sucks

Answer 71

yes but they don't introduce concept of derivative = slope in geometry I don't want to argue with you about this my point is simple and clear

Answer 72

You can have complex functions that you will have to find tangent lines to as well

Answer 73

Why would u want to do that?

Answer 74

I said I don't want to..please can you read...the question is solved so can we stop putting unnecessary notifications for the questioner?

Answer 75

Oooh, the questioner can say that themselves, no?

Answer 76

lol estudier why are you so against my method...please explain..I told you understand that its not useful only if the instructor did not inform them to use a calculus approach..that's all I'm saying

Answer 77

I am not against your method, if that is what floats your boat, go ahead.

All I'm saying is that it's a 1 liner, no calculus required.

Answer 78

If I were the instructor, I wouldn't care what method you use. Long as you get the correct answer and show your steps. I'd give extra credit to those who solve without calculus, lol

Answer 79

haha...good for you Hero

Answer 80

Nayeaddo. I'm taking precalc. This is actually part of my summer math from alg 2 lol

Answer 81

lol

Answer 82

and I still don't get it btw.

Answer 83

I can show you on twiddla

Answer 84

what's twiddla?

Answer 85

My way is the simplest

Answer 86

Just look at the link I posted, it's obvious...

Answer 87

One liner.

Answer 88

None of the lines on your link had a slope of -1 estudier

Answer 89

Estudier, even you have to admit it

Answer 90

What?

Answer 91

the line I'm looking for has a slope of -1 and is tangent to the circle

Answer 92

There are 2 such lines...

Answer 93

http://www.twiddla.com/solved

Answer 94

3rt2-x = -1?

Answer 95

The line y = -x...does not have a slope of -1?

Answer 96

Since when?

Answer 97

no such line 0.o?

Answer 98

oh