Java and Analytic geometry

[Haf912]

Hello there,

A group of my friends and I are doing a school project about recreating paint in Java. I wanted to add something unique so i decided that i'll a add the possibility to draw a circle that, given three points, touches them. Like so : 
 

 

I am encountering some problems, not from a logical point of view but when I need to "translate" math language into Java i don't now where to start.

Here's what I would do in math:

• Find the equation of the line that passes between A and B with the formula 
• Do the same for the line that passes between B and C using the same formula
• Find the "middle point" (I don't know if the term is correct) of AB and BC using the formula 
• I have the m of the equation y = mx +q so I can just obtain the m of the axis by doing the anti reciprocal of the m of the lines that passes between A and B, B and C
• Obtain the equation of the lines that passes between the middle point (axis) of AB and BC, knowing m using
• Do a system between the equation of the two axis and find the center of the circle
• Draw a circle

I only know that for resolving the system it is more convenient to use Cramer's method   but that's the only thing I know.

Any help will be appreciated. If something is not clear feel free to ask, especially if you don't understand something due to linguistic unclarity because I am not an English native speaker. Thanks

[Ciccioo]

you need to find the generic equation of the generic circle that passes through three generic points

the equation of a circle is in the form

(x - a)^2 + (y - b)^2 = r^2

so you need to find the three functions f, g, h so that given three points P1, P2, P3

a = f(P1, P2, P3)b = g(P1, P2, P3)r = h(P1, P2, P3)

at this point it's just a math problem

 

i think that a simpler approach to the problem is to only use the generic equation of a circumference instead of making segments, intersecting them and stuff.

you can insert the coordinates of each of the three points into the equation of the circumference, so that you get a linear system of 3 equations (one for each point) in 3 variables (a, b, r) and you only need to solve that

[Haf912]

you need to find the generic equation of the generic circle that passes through three generic points

the equation of a circle is in the form

(x - a)^2 + (y - b)^2 = r^2

so you need to find the three functions f, g, h so that given three points P1, P2, P3

a = f(P1, P2, P3)b = g(P1, P2, P3)r = h(P1, P2, P3)

at this point it's just a math problem

 

i think that a simpler approach to the problem is to only use the generic equation of a circumference instead of making segments, intersecting them and stuff.

you can insert the coordinates of each of the three points into the equation of the circumference, so that you get a linear system of 3 equations (one for each point) in 3 variables (a, b, r) and you only need to solve that

Yeah, you're right. Your method is much more simpler than mine.

Now the big question. How do i resolve three-way system in java ? Cramer's method ?

[Ciccioo]

Yeah, you're right. Your method is much more simpler than mine.

Now the big question. How do i resolve three-way system in java ? Cramer's method ?

it's much more simple than that: you need to work out the generic formula by hand, and you will end up with a formula that you can write down in java as a simple expression

 

so you have three points with generic coordinates:

P1(a, b )

P2(c, d)

P3(e, f)

plug those coordinates in the system and solve it using your favorite method