最优，，，

这个问题可以被建模为最小覆盖圆问题（Minimum Covering Circle Problem），其中平面被一个固定半径的圆覆盖。为了使移动的距离最短，我们希望找到一个最小半径的圆。

以下是用 MATLAB 实现 Welzl 算法解决最小覆盖圆问题的代码：

```matlab
function circle = welzl(points, boundary)
    if size(boundary, 1) == 3
        circle = circumcircle(boundary(1, :), boundary(2, :), boundary(3, :));
        return;
    elseif isempty(points) && size(boundary, 1) == 2
        circle = circumcenter(boundary(1, :), boundary(2, :));
        return;
    elseif isempty(points)
        circle = [];
        return;
    end

    p = points(end, :);
    points(end, :) = [];

    circle = welzl(points, boundary);
    if ~isempty(circle) && distance(circle.center, p) <= circle.radius
        return;
    end

    boundary = [boundary; p];
    circle = welzl(points, boundary);
    boundary(end, :) = [];

    return;
end

function d = distance(p1, p2)
    x1 = p1(1);
    y1 = p1(2);
    x2 = p2(1);
    y2 = p2(2);
    d = sqrt((x2 - x1)^2 + (y2 - y1)^2);
end

function circle = circumcircle(p1, p2, p3)
    x1 = p1(1);
    y1 = p1(2);
    x2 = p2(1);
    y2 = p2(2);
    x3 = p3(1);
    y3 = p3(2);
    
    d = 2 * (x1*(y2-y3) + x2*(y3-y1) + x3*(y1-y2));
    ux = ((x1^2 + y1^2)*(y2-y3) + (x2^2 + y2^2)*(y3-y1) + (x3^2 + y3^2)*(y1-y2)) / d;
    uy = ((x1^2 + y1^2)*(x3-x2) + (x2^2 + y2^2)*(x1-x3) + (x3^2 + y3^2)*(x2-x1)) / d;
    
    radius = distance([ux, uy], p1);
    circle.center = [ux, uy];
    circle.radius = radius;
end

function circle = circumcenter(p1, p2)
    x1 = p1(1);
    y1 = p1(2);
    x2 = p2(1);
    y2 = p2(2);
    
    ux = (x1 + x2) / 2;
    uy = (y1 + y2) / 2;
    
    radius = distance([ux, uy], p1);
    circle.center = [ux, uy];
    circle.radius = radius;
end

% 生成随机点集
points = rand(30, 2) * 10;

% 求解最小覆盖圆
boundary = [];
circle = welzl(points, boundary);

disp(['最小覆盖圆的中心：', num2str(circle.center)]);
disp(['最小覆盖圆的半径：', num2str(circle.radius)]);
```

在这个 MATLAB 代码示例中，我们生成了一个随机的点集，并通过调用 `welzl` 函数来解决最小覆盖圆问题。最后输出了最小覆盖圆的中心和半径。

你可以根据实际情况将代码适应到你的平面和覆盖面积尺寸上。