在最接近给定点的圆上找到点的最佳方法

Question

Given a point (pX, pY) and a circle with a known center (cX,cY) and radius (r), what is the shortest amount of code you can come up with to find the point on the circle closest to (pX, pY)?

I've got some code kind of working but it involves converting the circle to an equation of the form (x - cX)^2 + (y - cY)^2 = r^2 (where r is radius) and using the equation of the line from point (pX, pY) to (cX, cY) to create a quadratic equation to be solved.

Once I iron out the bugs it'll do, but it seems such an inelegant solution.

Answer 1

where P is the point, C is the center, and R is the radius, in a suitable "mathy" language:

V = (P - C); Answer = C + V / |V| * R;

where |V| is length of V.

OK, OK

double vX = pX - cX;
double vY = pY - cY;
double magV = sqrt(vX*vX + vY*vY);
double aX = cX + vX / magV * R;
double aY = cY + vY / magV * R;

easy to extend to >2 dimensions.

Answer 2

我会从中心到点画一条线，然后计算该图与圆的交点oO我认为并不那么困难

Answer 3

Solve it mathematically first, then translate into code. Remember that the shortest line between a point and the edge of a circle will also pass through its center (as stated by @litb).

Answer 4

1. The shortest distance point lies at the intersection of circumference and line passing through the center and the input point. Also center, input and output points lie on a straight line

2. let the center be (xc, yc) and shortest point from input (xi, yi) be (x,y) then sqrt((xc-x)^2 + (yc-y)^2) = r

3. since center, input and output points lie on a straight line, slope calculated between any of two of these points should be same.

(yc-yi)/(xc-xi) = (y-yc)/(x-xc)

4.solving equations 2&3 should give us the shortest point.

Answer 5

触发函数，乘以r，然后适当加上pX或pY。

Answer 6

将圆心作为原点，将（pX，pY）的坐标转换为极坐标，（theta，r'）将r'替换为原始圆的r，然后转换回直角坐标（并调整原点）。

Answer 7

You asked for the shortest code, so here it is. In four lines it can be done, although there is still a quadratic. I've considered the point to be outside the circle. I've not considered what happens if the point is directly above or below the circle center, that is cX=pX.

m=(cY-pY)/(cX-pX);  //slope
b=cY-m*cX;  //or Py-m*Px.  Now you have a line in the form y=m*x+b
X=(  (2mcY)*((-2*m*cY)^2-4*(cY^2+cX^2-b^2-2*b*cY-r^2)*(-1-m^2))^(1/2)  )/(2*(cY^2+cX^2-b^2-2*bc*Y-r^2));
Y=mX+b;

1] Get an equation for a line connecting the point and the circle center.

2] Move along the line a distance of one radius from the center to find the point on the circle. That is: radius=a^2+b^2 which is: r=((cY-Y)+(cX-X))^(1/2)

3] Solve quadratically. X=quadratic_solver(r=((cY-Y)+(cX-X))^(1/2),X) which if you substitute in Y=m*X+b you get that hell above.

4] X and Y are your results on the circle.

I am rather certain I have made an error somewhere, please leave a comment if anyone finds something. Of course it is degenerate, one answer is furthest from your point and the other is closest.

Answer 8

Easy way to think about it in terms of a picture, and easy to turn into code: Take the vector (pX - cX, pY - cY) from the center to the point. Divide by its length sqrt(blah blah blah), multiply by radius. Add this to (cX, cY).

Answer 9

Here is a simple method I use in unity... for the math kn00bs amongst us. Its dependent on the transform orientation but it works nicely. I am doing a postion.z = 0 but just fatten the axis of the 2d circle you are not using.

//Find closest point on circle
Vector3 closestPoint = transform.InverseTransformPoint(m_testPosition.position);
closestPoint.z = 0;
closestPoint = closestPoint.normalized * m_radius;

Gizmos.color = Color.yellow;
Gizmos.DrawWireSphere(transform.TransformPoint(closestPoint), 0.01f);