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

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SUMMARY

The discussion focuses on finding the equation of the tangent line to a circle centered at (3,2) with a tangent point at (8,4). The radius of the circle is calculated using the distance formula, yielding a radius of sqrt(29). The slope of the radius is determined to be 2/5, leading to the conclusion that the slope of the tangent line is -5/2, as the tangent is perpendicular to the radius. The final equation of the tangent line is expressed in point-slope form as y - 4 = -5/2(x - 8).

PREREQUISITES
  • Understanding of the distance formula for two points in a Cartesian plane.
  • Knowledge of the relationship between the slopes of perpendicular lines.
  • Familiarity with point-slope form of a linear equation.
  • Basic concepts of circles, including center and radius.
NEXT STEPS
  • Study the distance formula in detail to calculate distances between points.
  • Learn about the properties of perpendicular lines and their slopes.
  • Practice using point-slope form to write equations of lines.
  • Explore the general equation of a circle and its applications in geometry.
USEFUL FOR

Students studying geometry, particularly those focusing on circles and tangent lines, as well as educators looking for clear explanations of these concepts.

nukeman
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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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hi nukeman! :smile:

the tangent will be perpendicular to the radius :wink:
 
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
 
nukeman said:
… x intercept would be (9,0)

where do you get that from? :confused:

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

yup! :smile:

so the slope of the tangent is … ? :wink:​
 
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.
 
Oh ok, thanks Mark!

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