How to calculate the start point and the end point of a circle based on points - math

Question

I want bevel the sides of a rectangle in order to do so i want to draw a circle from the start point to the end point as shown in the image
while drawing a complete circle this is how i do it
float angle = 2.0f * M_PI * i / iSegments;
// vertex data
float x, y, z ,tx,ty ,tz;
x = cos(angle) * 50.0;
y = sin(angle) * 50.0;
z = 0.0;
How do we calculate the vertices for the circle in the given case ?

Answer 1

The center of the circle is radius away from the corner point, both in x and in y directions. In your example the radius is 30. To draw an arc of 90° (π/2 radians), you can start with the first angle and divide π/2 into as many segments as you want. The first angle depends on the which corner of the rectangle is treated. In the example of the upper right corner, the starting angle would be 0°.
Here is some code to illustrate the concept:
// draw a circular bevel to a rectangle, given are
// p_x, p_y: the coordinates of the corner, e.g. 50, 50
// rad: the radius of the bevel, e.g. 30
// dir_x: the direction to indicate whether the center of
// the circle lies lef (dir_y=-1) or right (dir_x=+1) of the corner
// dir_y: the direction to indicate whether the center of
// the circle lies lower (dir_y=-1) or higher (dir_y=+1) than the corner
void draw_circular_bevel (float p_x, float p_y, float rad, int dir_x, int dir_y)
{
float c_x, c_y; // the center of the circle
float start_angle; // the angle where to start the arc
c_x = p_x + rad * dir_x;
c_y = p_y + rad * dir_y;
if (dir_x == 1 and dir_y == 1)
start_angle = 0.0;
else if (dir_x == 1 and dir_y == -1)
start_angle = - M_PI * 0.5f;
else if (dir_x == -1 and dir_y == 1)
start_angle = M_PI * 0.5f;
else if (dir_x == -1 and dir_y == -1)
start_angle = - 2.0f * M_PI;
for (int i=0; i <= iSegments; ++i) {
float x, y;
float angle = start_angle + 0.5f * M_PI * i / (float)iSegments;
x = c_x + cos(angle) * rad;
y = c_y + sin(angle) * rad;
// here call code to draw a point or a segment at position x,y
}
}