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### Finding the Center of a Circle

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

Hi,

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,

Steve
```

```
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
cases.

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
have:

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,
then:

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!  http://mathforum.org/dr.math/
```

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

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!  http://mathforum.org/dr.math/

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

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