Equation of the Tangent Line - Help please.

[nukeman]

1. The problem statement, all variables and given/known data

Center of circle is: (3,2)
Tangent point: (8,4)

Question: What is the equation of the tangent line?


2. Relevant equations



3. The attempt at a solution

I am just not getting it.

So, would the radius be 5?

Now, would i go:

(x-3)^2 + (y-2)^2 = 5

? Any help would be great!!

[tiny-tim]

hi nukeman! :smile:

the tangent will be perpendicular to the radius :wink:

[nukeman]

hi nukeman! :smile:

the tangent will be perpendicular to the radius :wink:

ooook. So, where I do from there?

The line would have a point of (8,4), and x intercept would be (9,0)

would the slope be 2/5 ?

Man, im lost

[tiny-tim]

… x intercept would be (9,0)

where do you get that from? :confused:

what is the slope of the radius? :smile:

[nukeman]

Would the slope be 2/5 ?

[tiny-tim]

Would the slope be 2/5 ?

yup! :smile:

so the slope of the tangent is … ? :wink:

[Mark44]

1. The problem statement, all variables and given/known data

Center of circle is: (3,2)
Tangent point: (8,4)

Question: What is the equation of the tangent line?


2. Relevant equations



3. The attempt at a solution

I am just not getting it.

So, would the radius be 5?

Now, would i go:

(x-3)^2 + (y-2)^2 = 5

? Any help would be great!!

This problem can be done without having to find the equation of the circle or its radius, but to set the record straight, the radius of the circle is not 5. The formula for the distance between two points (x1, y1) and (x1, y1), is sqrt((x2 - x1)2 + (y2 - y1)2). Applying this formula gives you sqrt(29) for the radius of the circle.

The equation of this circle would then be (x-3)2 + (y-2)2 = (sqrt(29))2 = 29.

[nukeman]

Oh ok, thanks Mark!

So, that would give me the equation of the circle. How would I get the equation of the tangent line?

[phinds]

Oh ok, thanks Mark!

So, that would give me the equation of the circle. How would I get the equation of the tangent line?

what is the slope of the radius out to the point where the tangent line hits?

what's the realtionship between the slope of the radius and the slope of the tangent line?

what is one point on the tangent line?

[Mark44]

As I said, the equation of the circle doesn't really enter into the problem.

If you know the slope of a line and a point on it, you can use the point-slope form of the equation of a line, which is y - y0 = m(x - y0).

Once you get the slope of the radius between the circle center and the point (8, 4), you can find the slope of the tangent line. The tangent line and the radius will be perpendicular, so what does that say about their slopes?

[nukeman]

So, they would have the same slope then?

The question in my book asks for

*The Equation of the Tangent Line
*The Equation of the Circle, stated in polynomic form

[phinds]

So, they would have the same slope then?

The question in my book asks for

*The Equation of the Tangent Line
*The Equation of the Circle, stated in polynomic form

Nukeman, I do not mean any disrespect here, but if getting the equation of the line is not completely clear to you based on what everyone has been telling you, you really need to study the basics a bit more thoroughly.

[nukeman]

I got how the equation of the circle would be (x-3)2 + (y-2)2 = (sqrt(29))2 = 29

The answer I got the equation to the tangent line comes out to, in

y-4 = 2/5x - 8
y= 2/5x - 2

[Mark44]

I got how the equation of the circle would be (x-3)2 + (y-2)2 = (sqrt(29))2 = 29
This isn't relevant to the problem.


The answer I got the equation to the tangent line comes out to, in

y-4 = 2/5x - 8
y= 2/5x - 2
Nope, you're missing an important point. The slope of the radius between (3, 2) and (8, 4) - the point of tangency - is 2/5, but that isn't the slope of the tangent line. The tangent line is perpendicular to the radius, so the slope of the tangent line will be ???.

[nukeman]

But this is the equation to the circle correct? (x-3)2 + (y-2)2 = (sqrt(29))2 = 29

Ok....Is the slope then, -5/2 (minus 5 over 2) ?

[Mark44]

But this is the equation to the circle correct? (x-3)2 + (y-2)2 = (sqrt(29))2 = 29
Yes, but so what? The question isn't asking for the equation of the circle.


Ok....Is the slope then, -5/2 (minus 5 over 2) ?
YES

Now, find the equation of the tangent line. Forget the equation of the circle.

[nukeman]

so, its

y-4 = -5/2 (x-8) ?

or,

y= -5/2x -1/2

?

[tiny-tim]

hi nukeman! :smile:

(just got up :zzz: …)
so, its

y-4 = -5/2 (x-8) ?

yes :smile:
or,

y= -5/2x -1/2

erm :redface: …

nooo! :rolleyes:
The question isn't asking for the equation of the circle.

he changed the question, in post #11 (http://www.physicsforums.com/showpost.php?p=3318749&postcount=11) ! :biggrin: