Simultaneous Equations

[Rich2189]

Whats the best way to to solve two equations simultaneouly? The method i use at the minute is to store each of the coeficients separatly. Then work on them. One equation is linear the other isn't.

Rich

[VBAhack]

If one of the equations is linear, it seems to me substitution would work. What are the 2 equations?

[Rich2189]

y=X+1 and (X-3)^2+(Y-4)^2 = 4

Yeah substitution works fine but the code sucks, i was wondering if there was a better way than substitution.

Rich

[Andrew G]

You can graph both solutions then look for where they intersect, but i don't recomend that

[VBAhack]

Hmmmm, it's not clear what you are trying to do. No code is needed. This can readily be solved with algebra. The 2nd equation is a circle and the first is a line, so there are either no solutions (i.e. the line doesn't touch or cross the circle), one solution (i.e. line is tangent to the circle at one point), or 2 solutions (line crosses the circle at 2 points). In this case, there are 2 solutions. Using substitution, you get (x-3)^2 = 2, which yields x = 3 + sqrt(2) and x = 3 - sqrt(2).

What are you trying to do in code? Maybe you can post it?

[Rich2189]

This is my function, works splendind, but is there a better way to do it?

Public Function chordCircleIntersection(chordPoint1 As POINT, chordPoint2 As POINT, radius As Long, center As POINT) As ROOT
 
    
    Dim units2 As Long, m As Long, units As Long, Xco As Long, xSqrco As Long
    
        m = (chordPoint2.Y - chordPoint1.Y) / (chordPoint2.X - chordPoint1.X)
    
    
    
    units = -chordPoint1.X * m + (chordPoint1.Y)
    xSqrco = (m ^ 2) + 1
    Xco = (-chordPoint1.X * 2) + ((-chordPoint1.Y + units) * 2)
    units2 = (chordPoint1.X ^ 2) + (((-chordPoint1.Y) + units) ^ 2) + (-radius ^ 2)
 
 
    Dim discriminant As Long, root1 As Double, root2 As Double, minusB As Long
 
    minusB = Xco * -1
    discriminant = (Xco ^ 2) - (4 * xSqrco * units2)
 
    If discriminant >= 0 Then
        chordCircleIntersection.root1.X = (minusB + (discriminant ^ 0.5)) / (xSqrco * 2)
        chordCircleIntersection.root2.X = (minusB - (discriminant ^ 0.5)) / (xSqrco * 2)
        'caculate the Y coords
        
        chordCircleIntersection.root1.Y = (chordCircleIntersection.root1.X * m) + units
        chordCircleIntersection.root2.Y = (chordCircleIntersection.root2.X * m) + units
    Else
        MsgBox "The Chord Doesn't Pass through the circle" & vbNewLine & "chordCircleIntersection", vbCritical, "Organic Molecule"
    End If
 
 
 
End Function

Rich

[VBAhack]

Yeah substitution works fine but the code sucks, i was wondering if there was a better way than substitution

What sucks about it? Your function is a generalized one to find the intersection of a line and a circle. From the standpoint of the mathematical approach, I think what you have is about as simple as it gets. Quite frankly, I can't think of any other way to do it, except maybe some sort of iterative technique.

If you are looking for a critique of coding style/elegance/efficiency (e.g. using UDT's as return values, etc.), other (more experienced coders) would need to reply.

[Rich2189]

What sucks about it? Your function is a generalized one to find the intersection of a line and a circle. From the standpoint of the mathematical approach, I think what you have is about as simple as it gets. Quite frankly, I can't think of any other way to do it, except maybe some sort of iterative technique.

If you are looking for a critique of coding style/elegance/efficiency (e.g. using UDT's as return values, etc.), other (more experienced coders) would need to reply.

Thanks that was what i was looking for, i just supposed their must be a better way to find those points Aparently not. . the thing that sucks is simulating the algebra, boggles the mind and very hard to debug.