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By : tuschenski
Date : November 22 2020, 10:40 AM
I hope this helps you . What you need is a new coordinate system to place the circle. As any common coordinate system, we'll want the base vectors to be orthogonal to each other, and have length 1 each. I'll name the base vectors v1, v2, and v3, which correspond to x, y, and z in order.
The new base vector that replaces z, which is v3 is given by the normal vector of the circle. If it's not normalized yet, you'll want to normalize it here: code :
``````     [ v3x ]
v3 = [ v3y ] = normalize(circleNormal)
[ v3z ]
``````
``````               [ v3z ]
v1 = normalize([ 0   ])
[ -v3x]
``````
``````v2 = v3 x v1
``````
``````p = centerPoint + R * (cos(a) * v1 + sin(a) * v2)
``````
``````// Only needed if normal vector (nx, ny, nz) is not already normalized.
float s = 1.0f / (nx * nx + ny * ny + nz * nz);
float v3x = s * nx;
float v3y = s * ny;
float v3z = s * nz;

// Calculate v1.
s = 1.0f / (v3x * v3x + v3z * v3z);
float v1x = s * v3z;
float v1y = 0.0f;
float v1z = s * -v3x;

// Calculate v2 as cross product of v3 and v1.
// Since v1y is 0, it could be removed from the following calculations. Keeping it for consistency.
float v2x = v3y * v1z - v3z * v1y;
float v2y = v3z * v1x - v3x * v1z;
float v2z = v3x * v1y - v3y * v1x;

// For each circle point.
px = cx + r * (v1x * cos(a) + v2x * sin(a))
py = cy + r * (v1y * cos(a) + v2y * sin(a))
pz = cz + r * (v1z * cos(a) + v2z * sin(a))
`````` ## Find the point on a circle with given center point, radius, and degree

By : Gustsaiyonee
Date : March 29 2020, 07:55 AM
To fix this issue The simple equations from your link give the X and Y coordinates of the point on the circle relative to the center of the circle.
code :
``````X = r * cosine(angle)
Y = r * sine(angle)
``````
``````X = Cx + (r * cosine(angle))
Y = Cy + (r * sine(angle))
`````` ## How to calculate a point on a circle knowing the radius and center point

By : Tamil
Date : March 29 2020, 07:55 AM
it should still fix some issue As you build up the track, the position of the next piece of track to be placed needs to be relative to location and direction of the current end of the track.
I would store an (x,y) position and an angle a to indicate the current point (with x,y starting at 0, and a starting at pi/2 radians, which corresponds to straight up in the "anticlockwise from 3-o'clock" system).
code :
``````fx = cos(a);
fy = sin(a);
lx = -sin(a);
ly = cos(a);
``````
``````x=x+112*fx
y=y+112*fy
``````
``````x=x+285.206*sin(22.5*pi/180)*fx // Move forward
y=y+285.206*sin(22.5*pi/180)*fy

x=x+285.206*(1-cos(22.5*pi/180))*(-lx) // Side-step right
y=y+285.206*(1-cos(22.5*pi/180))*(-ly)

a=a-22.5*pi/180 // Turn to face new direction
`````` ## Formula to find points on the circumference of a circle, given the center of the circle and the radius

By : onopuccino
Date : March 29 2020, 07:55 AM
I wish did fix the issue. I am working on code to find the points on the circumference of a circle. I have the center point of the circle and the radius and I need to draw a circle around it. This will help me define the boundary. Please help me with formula for finding the these points on the circumference. , For a circle with origin (j, k) and radius r:
x(t) = r cos(t) + j ## Javascript: compute the radius of circle given center point and another point

Date : March 29 2020, 07:55 AM
may help you . Suppose I have a circle with a center point at origin (0,0) and a 90deg point from it at (0,10)... from this 2 obvious points the radius of the circle would definitely be 10, right? , Power in javascript is done by using Math.pow:
code :
``````Math.sqrt( Math.pow((x1-x2), 2) + Math.pow((y1-y2), 2) )
`````` ## How to move n degrees from vector point A along the circumference of a circle? (image included)

By : Godwin
Date : March 29 2020, 07:55 AM
Hope that helps A, B and Center are 2D vector points. , All angles are in radians. Your n is whats called the circles arc. 