Finding the Center of a CircleDate: 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 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! http://mathforum.org/dr.math/ |
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