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Finding the Center of a Circle

Date: 12/26/96 at 20:12:45
From: Steve Heikke
Subject: Analytic geometry


My name is Steve and I am a machinist with a problem that I hope you 
can help me with.  I have been in the trade for roughly 35 years now 
and this problem comes up from time to time:
Given:  1. A circle of radius R = 0.02 with center point unknown
        2. A line of equation Y = mx+b (m is the tangent of 50                                      
           degrees, and b is - .1)                  
        3. A line at Y = -.08

The problem is to generate the x,y coordinates of the points of 
tangency where the two lines intersect with the circle.  I believe I 
could solve this problem myself if I had a way of determining the 
center point of the circle.  I have gone round and round with this 
problem and am at my wits end.  I would much appreciate any help you 
could send my way.
Thank you very much,

Date: 12/27/96 at 10:15:34
From: Doctor Jerry
Subject: Re: Analytic geometry

Hi Steve,

My son is a machinist and often gives me problems of this kind. 

I'm not certain where the circle is, but from your description of the 
lines, I'm assuming that it is pushed into one of the four angles 
formed by the two lines.  I don't think it matters which angle they 
form; the idea of how to find the center will be the same in all 

Based on my understanding of the problem, here is one method of 
solving it.  The method is based on a formula for the distance d from 
a point (p,q) to a line A*x+B*y+C=0. The formula is:

     d = |A*p+B*q+C|/sqrt(A^2+B^2)

I've used the absolute value signs |...| in this formula.  The 
distance d is the shortest distance from the point to the line, that 
is, it is the "perpendicular distance."  Suppose, for example, that 
the line is y = tan(50)x - 0.1 and the point is (0.3,0.2).  To 
calculate the distance d from the point to the line, first write the 
line in the form A*x+B*y+C=0.  This would be:

tan(50)x + (-1)y + (-0.1) = 0

From the formula, the distance d would be:

d = |tan(50)*0.3 + (-1)*0.2 + (-0.1)|/sqrt((tan(50))^2 + (-1)^2)

d = 0.057526/1.555724 = 0.036977

Now, suppose the circle lies in the upper right angle formed by the 
two lines you gave. Let (h,k) be the (unknown) coordinates of the 
center of the circle. Since one of the lines is y = -0.08 and the 
circle is tangent to this line, k = -0.1+0.02 = -0.08. We know that 
the distance from the center (h,k) of the circle to either of the 
lines is equal to the radius of the circle, that is, d = 0.02. Using 
the point (h,k) and the line with equation y = tan(50)x - 0.1, we 

d = 0.02 = |tan(50)*h + (-1)*k + (-0.1)|/sqrt((tan(50))^2 + (-1)^2)

    0.02 = |(1.191754*h + 0.08 - 0.1]/1.555724

    0.02*1.555724 = |(1.191754*h - 0.02|

There are two values of h that satisfy this equality. If you make a 
rough sketch, it looks as one of them is approximately 0.04. If so, 
the expression 1.191754*h - 0.02 is positive and we may remove the 
absolute value signs with no other change.  This gives:

0.031114 = 1.191754*h - 0.02 

0.051114 = 1.191754*h

h = 0.042890

If we assume that h makes the expression 1.191754*h - 0.02 negative, 

0.031114 = -(1.191754*h - 0.02) 

and h = -0.00932615

The other two solutions can be found with this procedure by taking 
k = -0.08 - 0.02 = -0.10 and doing a similar calculation.

Please let me know if this is understandable.

-Doctor Jerry,  The Math Forum
 Check out our web site!   

Date: 01/01/97 at 13:52:31
From: Steve Heikke
Subject: Re: Analytic geometry

Your swift reply took me by surprise. I had to read through your 
explanation several times before it sank in (I haven't had a math 
class in over 25 years, so am unfamiliar with the notation ).  I was 
then able to solve my problem with only a mild headache as a result.  
I never knew that there was an equation for the distance from a point 
to a line.  This knowledge will help me in the future.  I have 
transcribed the equation and put it into a little black book in my 
toolbox at work.

By way of background information for you, I work as a machinist for 
the Boeing Company in Everett, WA.  I was able to complete the second 
quarter of calculus 25 years ago but then had to drop out.  Many times 
I have thought about going back, just to see how far I could go the 
second time around.  Unfortunatly there are other demands on my time 
(there's always an excuse) and I fear that I never will.  There are
a lot of things to learn about in this life and I believe I know 
90 percent of the mathematics that I need for my career.
Again thank you very much for your prompt reply to my question.

Happy New Year (1997),
Steve Heikke

Date: 01/01/97 at 15:59:58
From: Doctor Jerry
Subject: Re: Analytic geometry

Thanks, Steve, and a happy new year to you too.  By the way, I taught 
calculus at Iowa State Univ for 35 years or so.

-Doctor Jerry,  The Math Forum
 Check out our web site!   

Associated Topics:
College Euclidean Geometry
High School Euclidean/Plane Geometry

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