What is the angular coefficient of the tangent line to two circumferences?

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Homework Statement



Find the angular coefficient of the line that is tangent to the following circumferences:
[tex](x - 17)^{2} + y^{2} = 16[/tex]
[tex]x^{2} + y^{2} = 16[/tex]

Homework Equations




The Attempt at a Solution



I tried everything but nothing is working, please help me.
 
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Taturana said:

Homework Statement



Find the angular coefficient of the line that is tangent to the following circumferences:
[tex](x - 17)^{2} + y^{2} = 16[/tex]
[tex]x^{2} + y^{2} = 16[/tex]

Homework Equations




The Attempt at a Solution



I tried everything but nothing is working, please help me.
What did you try? Have you drawn a picture of the two circles (not circumferences)?

By "angular coefficient" do you mean slope?
 
Mark44 said:
What did you try? Have you drawn a picture of the two circles (not circumferences)?

By "angular coefficient" do you mean slope?

Yeh I mean slope.

here is a picture:

[PLAIN]http://img94.imageshack.us/img94/2910/58393508.png

I tried to make a system of equations such that: the distance between the center of the circumferences and the line is equal to 4 (that is the radius of the circumferences). But I end up to something like |17a + c| = |c| (considering the line as ax + by + c = 0), but it doesn't help me.

I tried to make two systems:
1: using the equation of the first circumference and the equation of the line
2: using the equation of the second circumference and the equation of the line
then I shared some variables between these systems (a and b, considering y = ax + b as the line). But the equations become very complicated and I think it's not the easiest way.

Someone can help me?
 
Last edited by a moderator:
Writing equations probably isn't the easiest way to solve it. Why don't you draw some right triangles in your picture?
 
Dick said:
Writing equations probably isn't the easiest way to solve it. Why don't you draw some right triangles in your picture?

Thank you, you helped me a lot, now I solved.