×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

# A challenge for Mathcad Prime 5.0

## A challenge for Mathcad Prime 5.0

(OP)
I am still waiting for PTC to get their act together. I have 25 years of Mathcad files.

I stopped upgrading at Mathcad 13. I use the symbolic solver a lot and PTC ruined it when they made Mathcad 14. Now I use Mathematica that is expensive. I also use Python with sympy to do symbolic processing.
So here is the challenge. Given 3 points solve for the center of the circle and radius to the 3 points.

This problem is simple enough to be entered in a few minutes.

#### CODE

from sympy import symbols, init_printing, sqrt
from sympy.solvers import solve

""" find the center and radius of a circle defined by three points """
init_printing()                         # enable pretty printing.
x0, y0, x1, y1, x2, y2, xc, yc, r = symbols('x0, y0, x1, y1, x2, y2, xc, yc, r')
eq0 = (x0-xc)**2 + (y0-yc)**2 - r**2
eq1 = (x1-xc)**2 + (y1-yc)**2 - r**2
eq2 = (x2-xc)**2 + (y2-yc)**2 - r**2
s = solve([eq0, eq1, eq2], (xc, yc, r))
# there are two solutions, r > 0 and r < 0
xc0, yc0, r0 = s[0]
print("\nxc0 =",xc0)
print("\nyc0 =",yc0)
print("\nr0  =",r0)
xc1, yc1, r1 = s[1]
print("\nxc1 =",xc1)
print("\nyc1 =",yc1)
print("\nr1  =",r1) 

Here is the output. There are two solutions because r can be positive or negative and still be valid as far as the problem goes but only the positive solution works.

#### CODE

xc0 = (x0**2*y1 - x0**2*y2 - x1**2*y0 + x1**2*y2 + x2**2*y0 - x2**2*y1 + y0**2*y1 - y0**2*y2 - y0*y1**2 + y0*y2**2 + y1**2*y2 - y1*y2**2)/(2*(x0*y1 - x0*y2 - x1*y0 + x1*y2 + x2*y0 - x2*y1))

yc0 = (-x0**2*x1 + x0**2*x2 + x0*x1**2 - x0*x2**2 + x0*y1**2 - x0*y2**2 - x1**2*x2 + x1*x2**2 - x1*y0**2 + x1*y2**2 + x2*y0**2 - x2*y1**2)/(2*(x0*y1 - x0*y2 - x1*y0 + x1*y2 + x2*y0 - x2*y1))

r0  = sqrt((x0**2 - 2*x0*x1 + x1**2 + y0**2 - 2*y0*y1 + y1**2)*(x0**2 - 2*x0*x2 + x2**2 + y0**2 - 2*y0*y2 + y2**2)*(x1**2 - 2*x1*x2 + x2**2 + y1**2 - 2*y1*y2 + y2**2))/(2*(x0*y1 - x0*y2 - x1*y0 + x1*y2 + x2*y0 - x2*y1))

xc1 = (x0**2*y1 - x0**2*y2 - x1**2*y0 + x1**2*y2 + x2**2*y0 - x2**2*y1 + y0**2*y1 - y0**2*y2 - y0*y1**2 + y0*y2**2 + y1**2*y2 - y1*y2**2)/(2*(x0*y1 - x0*y2 - x1*y0 + x1*y2 + x2*y0 - x2*y1))

yc1 = (-x0**2*x1 + x0**2*x2 + x0*x1**2 - x0*x2**2 + x0*y1**2 - x0*y2**2 - x1**2*x2 + x1*x2**2 - x1*y0**2 + x1*y2**2 + x2*y0**2 - x2*y1**2)/(2*(x0*y1 - x0*y2 - x1*y0 + x1*y2 + x2*y0 - x2*y1))

r1  = -sqrt((x0**2 - 2*x0*x1 + x1**2 + y0**2 - 2*y0*y1 + y1**2)*(x0**2 - 2*x0*x2 + x2**2 + y0**2 - 2*y0*y2 + y2**2)*(x1**2 - 2*x1*x2 + x2**2 + y1**2 - 2*y1*y2 + y2**2))/(2*x0*y1 - 2*x0*y2 - 2*x1*y0 + 2*x1*y2 + 2*x2*y0 - 2*x2*y1) 
The generated code is much more efficient that Mathcad 13. Notice I can use text which is very convenient for educational purpose or solving math for a webpage with a java script interface. Python will also generate LaTex for pretty printing on web pages.

What can Mathcad Prime 5.0 do?

Peter Nachtwey
Delta Computer Systems
http://www.deltamotion.com
http://forum.deltamotion.com/

#### Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

#### Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!