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About Circle


Sriman Dutta

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Hello,

Suppose there's a circle with radius r and centre C. Let there be a point outside the circle P, such that the length PC is known. Now, lets take a point X on the circumference of the circle, such that the inclination of CX with respect to PC is a. Is it possible to find PX? If so, how??

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By inclination do you mean the tangent of the angle XCP? If so you have side-angle-side and you can determine the third side PX. https://www.mathsisfun.com/algebra/trig-solving-sas-triangles.html

 

In other words we know PC, and we know that CX = r, and we know the angle XCP as the arctangent of the inclination.

 

Am I understanding inclination correctly? As the slope of the line with respect to CP, in other words imagining CP as the positive x-axis and then the inclination is the slope of CX?

Edited by wtf
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There are four possibilities conforming to your description.

 

For the first 3 wtf has told you how to find PX by solving the triangle PCX.

Note that PX does not need to be a tangent it may cut the circle.

 

In the fourth possibility PCX are colinear so PX = PC + r and a = 180o

 

Note that for

 

case 1 the triangle is a right angled triangle so a simple formula is used

 

case 2 and case 3 uses the cosine rule.

Note that for acute angles ( a<90) cos(a) is edit (oops :eek: beware the double negative) negative positive so the last term is negative, but for obtuse angles (a>90) cos(a) is negative so the last term is positive.

Note also that since we are using the cosine rule there are no positions of P that cannot be solved, unlike for a sine rule problem.

 

case 4 is again a simpler equation.

 

post-74263-0-83954500-1475053521_thumb.jpg

Edited by studiot
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Note that for acute angles ...( a<90)

A small quibble. OP said that a is the inclination, not the angle. The inclination is the slope of CX with respect to PC.

 

In other words the angle is the arctan of a. OP did not provide a clarification so I assume this is what is meant.

 

(In your pictures this would be more clear if you drew CX as having a positive angle with PC. In your pictures, the inclination is negative.)

Edited by wtf
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A small quibble. OP said that a is the inclination, not the angle. The inclination is the slope of CX with respect to PC.

 

In other words the angle is the arctan of a. OP did not provide a clarification so I assume this is what is meant.

 

(In your pictures this would be more clear if you drew CX as having a positive angle with PC. In your pictures, the inclination is negative.)

 

I don't suppose your visits overlapped.

 

I think formal study is difficult in India and the OP is trying his best with the English language, so I took it to mean to angle

 

I'm glad it was a small quibble since inclinometers (I have several) measure angle.

 

https://en.wikipedia.org/wiki/Inclinometer

 

They were much used in the survey of India.

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I don't suppose your visits overlapped.

 

I think formal study is difficult in India and the OP is trying his best with the English language, so I took it to mean to angle

 

I'm glad it was a small quibble since inclinometers (I have several) measure angle.

 

https://en.wikipedia.org/wiki/Inclinometer

 

They were much used in the survey of India.

You are right. I looked up inclination last night and for some reason I thought it said the slope, but actually it said the angle. My mistake. Inclination is the angle.

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  • 2 months later...

Are there not two colinear solutions?

 

[math] a=180^o [/math]

 

and

 

[math] a=0^o [/math]

 

Yes but a = 0 is the limiting situation for my case 1, with a=0, PX = PCsin(a) = PCsin(0) = 0.

 

Does this calculation work for a = 180? (case 4)

 

:)

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