How Do You Find the Equation of a Tangent Line to a Circle?

In summary, the equation of the tangent line to a circle at the point (8,4) with a center at (3,2) is y-4 = -5/2 (x-8). The slope of the tangent line is -5/2 and it is perpendicular to the radius of the circle passing through the center and the tangent point. The equation of the circle itself is (x-3)^2 + (y-2)^2 = 29.
  • #1
nukeman
655
0

Homework Statement



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

Question: What is the equation of the tangent line?


Homework Equations





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!
 
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  • #2
hi nukeman! :smile:

the tangent will be perpendicular to the radius :wink:
 
  • #3
tiny-tim said:
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, I am lost
 
  • #4
nukeman said:
… x intercept would be (9,0)

where do you get that from? :confused:

what is the slope of the radius? :smile:​
 
  • #5
Would the slope be 2/5 ?
 
  • #6
nukeman said:
Would the slope be 2/5 ?

yup! :smile:

so the slope of the tangent is … ? :wink:​
 
  • #7
nukeman said:

Homework Statement



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

Question: What is the equation of the tangent line?


Homework Equations





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.
 
  • #8
Oh ok, thanks Mark!

So, that would give me the equation of the circle. How would I get the equation of the tangent line?
 
  • #9
nukeman said:
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?
 
  • #10
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?
 
  • #11
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
 
  • #12
nukeman said:
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.
 
  • #13
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
 
  • #14
nukeman said:
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.
nukeman said:
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 ?.
 
  • #15
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) ?
 
  • #16
nukeman said:
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.
nukeman said:
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.
 
  • #17
so, its

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

or,

y= -5/2x -1/2

?
 
  • #18
Last edited by a moderator:

1. What is the equation of a tangent line?

The equation of a tangent line is a linear equation that represents the slope of a curve at a specific point. It is used to find the rate of change or slope of a curve at a single point.

2. How do you find the equation of a tangent line?

The equation of a tangent line can be found using the point-slope form of a line. First, find the slope of the curve at the given point using the derivative. Then, plug in the given point and slope into the point-slope form to get the equation of the tangent line.

3. What is the difference between a tangent line and a secant line?

A tangent line touches a curve at a single point, while a secant line intersects the curve at two points. The slope of a tangent line is the instantaneous rate of change at that specific point, while the slope of a secant line is the average rate of change between the two points.

4. Can you use the equation of the tangent line to find the derivative of a function?

Yes, the equation of the tangent line can be used to find the derivative of a function at a specific point. The slope of the tangent line is the same as the slope of the curve at that point, which is the definition of the derivative.

5. Is the equation of the tangent line the same as the derivative of a function?

No, the equation of the tangent line is a linear equation that represents the slope of a curve at a specific point. The derivative of a function is a general formula that represents the slope of a curve at any point. However, the equation of the tangent line can be used to find the derivative at a specific point.

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