Math-to-code problem

[GMMorris]

Hi,

 

I am trying to, programaticaly, calculate two points in my form, and although I am very easily capable of calculating these points on paper, I can't find a way to transfer the calculations into code, perhaps you can help.

:cool:

 

To the details -

 

I have two imaginary cicrles, located on my form (the specifics of why, are not at my disposale to reveal).

 

I know the center locations of the two circles, and I know the radius.

 

What I need is the (X,Y) locations of the two nearest points oin the two circles.

 

What I mean is - two points, one on each circle, between which the shortest line can be drawn.

 

Conclusions my calculations brought me to:

1)The two points will sit on the two points where the line running between the two circle's centers hits the actual circles.

2)The length of the shortest line = the distance between the two centers minus (-) twice the radius.

 

Look at the picture for extra clarification.

 

Please help me - this is very important

[Iceplug]

OK.

Radius1 is upper left radius, Radius2 is bottom right radius

First of all, you can determine the length of the line like this:

Length = Math.Sqrt( (J - M)^2 + (K - N)^2 )

 

Next, you can find the slope of the line like this:

(It is important that the second part is subtracted from the first or vice versa consistently.)

dY = (K - N) / Length

dX = (J - M) / Length

m = dY/dX

 

Then, you can find the first point like this:

A = J + dX * Radius1

B = K + dY * Radius1

 

Finally, the second point:

C = M - dX * Radius2

D = N - dY * Radius2

 

Hope that helps. :)

 

You'll probably have to declare most of those variables as singles. (esp. dY, dX, and m, if you need it)

[GMMorris]

how do you figure that?

 

Then, you can find the first point like this:

A = J + dX * Radius1

B = K + dY * Radius1

 

Finally, the second point:

C = M - dX * Radius2

D = N - dY * Radius2

 

oh, and if this is correct - you are my king :)

[Iceplug]

dX and dY form a unit vector for the line... if a line is flat on the x-axis, dX = 1, dY = 0

if a line is standing on y-axis, dX = 0, dY = 1

if it is leaning at 45 degrees, dX = 0.707 (1 / Sqrt(2)), dY = 0.707

So, whatever you multiply to both dX and dY will give you a point along this line at whatever distance you have multiplied by.

Let me know if you want more detail. :)

[GMMorris]

Well, its 2 AM here, and my brain is begining to make clack-clack sounds, so I'll just save my self the headache and assume your right :) thanks.