interpolation - Algorithm to iteratively discover points on an arc described by three points -

i writing graphics application needs calculate , display list of points along curve arc described 3 points.

lets have points (1,1), (2,4) , (5,2). need algorithm can give me values of y each x 1 5 fall on interpolated arc.

i'm sure simple task math whizes out there, me it's bit beyond mathematical payscale.

thanks in advance!

so problem how compute center c = (c1, c2) , radius r of circumference given 3 points p = (p1, p2), q = (q1, q2) , s = (s1, s2).

the idea simple. consists in realizing that, definition, center has same distance 3 points p, q , s.

now, set of points equidistant pand q perpendicular segment pq incident @ mid point (p+q)/2. similarly, set of points equidistant q , s perpendicular qs passing thru (q+s)/2. so, center c must intersection of these 2 lines.

let's compute parametric equations of these 2 straight lines.

for need 2 additional functions call dist(a,b) computes distance between points a , b , perp(a,b) normalizes vector b-a dividing length (or norm) , answers perpendicular vector normalized vector (keep in mind perpendicular (a,b) (-b,a) because inner product 0)

dist((a1,a2),(b1,b2))     return sqrt(square(b1-a1) + square(b2-a2))  perp((a1,a2),(b1,b2))     dist := dist((a1,a2),(b1,b2)).     := (b1-a1)/dist.     b := (b2-a2)/dist.     return (-b,a). 

we can write parametric expressions of our 2 lines

(p+q)/2 + perp(p,q)*t (q+s)/2 + perp(q,s)*u 

note both parameters different, hence introduction of 2 variables t , u.

equating these parametric expressions:

(p+q)/2 + perp(p,q)*t = (q+s)/2 + perp(q,s)*u 

which consists of 2 linear equations, 1 each coordinate, , 2 unknowns t , u (see below). solution of 2x2 system gives values of parameters t , u injected parametric expressions give center c of circumference.

once c known, radius r can calculated r := dist(p,c).

---

linear equations

(p+q)/2 + perp(p,q)*t = (q+s)/2 + perp(q,s)*u 

first linear equation (coordinate x)

(p1+q1)/2 + (p2-q2)/dist(p,q)*t = (q1+s1)/2 + (q2-s2)/dist(q,s)*u 

second linear equation (coordinate y)

(p2+q2)/2 + (q1-p1)/dist(p,q)*t = (q2+s2)/2 + (s1-q1)/dist(q,s)*u 

linear system (2x2)

(p2-q2)/dist(p,q)*t + (s2-q2)/dist(q,s)*u = (s1-p1)/2 (q1-p1)/dist(p,q)*t + (q1-s1)/dist(q,s)*u = (s2-p2)/2