Find Equation of Circle Tangent to Line, X-Axis: (4,0)

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In summary, the conversation is about finding the equation of a circle that is tangent to both the x-axis at point (4,0) and the line 3x-4y-17=0. The directed distance and point slope formula are used to determine the equation, but the absolute value signs need to be eliminated. It is suggested to assume the center of the circle is at (4,k) and find the distance from the line to (4,k) to be equal to the radius of the circle. It is also mentioned that there may be two solutions, one in the third quadrant and one in the first quadrant.
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Find the equation of a circle which is tangent to the x-axis at point (4, 0) and is tangent to line 3x-4y-17=0

Directed Distance, point slope formula...

Ok so I knew that the x-axis is the equation y=0. I think but i am not sure that the directed distance between the tangent line 3x-4y-17=0 is equal to the distance form (4,0) to the center. So this is what you get...

Abs(3x-4y-17)/Square root(9+4) = Square root ((x-4)^2+(y)^2)

But i am not sure how to get rid of the absolute value signs and i am not sure if this will get me anywhere. Will it get me to a dead end? If so, what do i do?
 
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  • #2
it is tangent to the line 3x-4y-17=0 at what point? is it not given?
 
  • #3
Number1 said:
Find the equation of a circle which is tangent to the x-axis at point (4, 0) and is tangent to line 3x-4y-17=0

Directed Distance, point slope formula...

Ok so I knew that the x-axis is the equation y=0. I think but i am not sure that the directed distance between the tangent line 3x-4y-17=0 is equal to the distance form (4,0) to the center. So this is what you get...

Abs(3x-4y-17)/Square root(9+4) = Square root ((x-4)^2+(y)^2)

But i am not sure how to get rid of the absolute value signs and i am not sure if this will get me anywhere. Will it get me to a dead end? If so, what do i do?

You're assuming that the center of the circle is at (4, 0), which is not true. Assume that the center is at (4, k). You want the distance from the line to (4, k) to be equal to k, the radius of the circle.

If you square |x|, you get x2.
 
  • #4
I still don't understand. How do you know that the x value for the center is 4?? the x- value can be anything.
 
  • #5
The distance from the center to point (4,0) must be the same as the distance from the center to the tangent line.

You know that the value of x coordiante for the center must be 4 because the circle is tangent to y=0 (the x-axis) at that point! Try to sketch it.


Knowing the slope of the tangent line let's you know the angle at which it touches the circle, and from here you can find the rest
 
  • #6
Yea but if you try it out you get 2 circles! How do i know which one you are talking about? One is small another is large. The smaller one is on the bottom and the larger one is on the top.
 
  • #7
I haven't worked this problem, so it's possible that there are two circles. Can you be clearer when you say that one is "on the bottom" and one is "on the top"? On the bottom or top of what?
 
  • #8
I mean on the top of the x-axis or y=0 intercept and the bottom of the axis
 
  • #9
OK, that makes sense. The great big circle is tangent to the line at some point in the third quadrant, while the small circle is tangent in the first quadrant. Since the radii of the circles are different, you have two solutions. That's not a problem.
 
  • #10
Number1 said:
Yea but if you try it out you get 2 circles!

I didn't notice this till after drawing a diagram. The trick is that for the first big circle you'll have the equation in this form: [itex](x-4)^2+(y-k)^2=k^2[/itex] and then to find k you find the perpendicular distance from the line to the point (4,k).

For the second circle notice that while the radius is (say, r) it is in the 3rd quadrant so you'll have the circle equation in the form [itex](x-4)^2+(y+r)^2=r^2[/itex].
Notice how it is y+r because the r is going to be positive (since it's a distance) but the y coordinate of the centre of the circle is negative. Now you find the distance between the line and the point (4,-r).
 

FAQ: Find Equation of Circle Tangent to Line, X-Axis: (4,0)

What is the equation of a circle tangent to the X-axis at the point (4,0)?

The equation of a circle tangent to the X-axis at the point (4,0) can be expressed as (x-4)^2 + y^2 = r^2, where r is the radius of the circle.

How do you find the center of the circle tangent to the X-axis at (4,0)?

The center of the circle can be found by using the formula (h,k), where h is the x-coordinate and k is the y-coordinate. In this case, the center of the circle would be (4,0).

Can you find the radius of the circle without knowing the equation of the line?

Yes, the radius of the circle can be found by using the distance formula between the center of the circle and the point of tangency (4,0). The distance formula is d = √((x2-x1)^2 + (y2-y1)^2), where (x1,y1) is the center of the circle and (x2,y2) is the point of tangency.

How many circles are tangent to the X-axis at (4,0)?

There are infinitely many circles that can be tangent to the X-axis at (4,0). This is because the center of the circle can lie anywhere on a line parallel to the X-axis passing through the point (4,0), and the radius can vary.

Is there a specific method to find the equation of a circle tangent to a line?

Yes, there is a specific method to find the equation of a circle tangent to a line. This involves finding the distance between the center of the circle and the line, and then using this distance to find the radius of the circle. The center of the circle can then be found by using the distance and the slope of the line. Finally, the equation of the circle can be written in the form (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.

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