blocks|key|3959850|text|如果你正在处理圆，获得交集的正确方法是使用一些代数。有四种可能的情况:没有交集，一个交集(相切)，两个交集，和无限的交集(它们是同一个圆).Let我们专注于两个交集的情况。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3959851|从https://math.stackexchange.com/a/256123/647423中，您可以获得一个线性方程，该方程沿着通过两个交点的直线将x与y联系起来：|offset|length|3959852|−2x(x1center−x2center)−2y(y1center−y2center)+=+(r1)%5E2−(r2)%5E2−((x1center)%5E2−(x2center)%5E2)−((y1center)%5E2−(y2center)%5E2).|code-block|syntax|javascript|3959853|从这里你可以得到一个关于x的y的公式，然后将y代入你的一个圆公式中以获得x的二次方。如果你不想实现一个二次方程求解器，你可以像这样使用numpy.roots：|3959854|root_array+=+np.roots(quadratic_coeff,+linear_coeff,+constant_coef)|3959855|entityMap|0|LINK|mutability|MUTABLE|url|https://math.stackexchange.com/a/256123/647423|1|https://docs.scipy.org/doc/numpy/reference/generated/numpy.roots.html#numpy-roots^0|0|1|1A|0|0|0|1V|B|1|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@$D|10|E|11|1|12]]|A|$]]|$1|F|3|G|5|H|7|13|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|14|8|@]|9|@$D|15|E|16|1|17]]|A|$]]|$1|M|3|N|5|H|7|18|8|@]|9|@]|A|$I|J]]|$1|O|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]|W|$5|R|S|T|A|$U|X]]]]

If you are working with circles, the proper approach to get the intersections is to use some algebra. There are four possible cases: no intersection, one intersection (tangency), two intersections, and infinite intersection (they are the same circle).Let us focus on the case of two intersections.

From <a href="https://math.stackexchange.com/a/256123/647423">https://math.stackexchange.com/a/256123/647423</a> what you can do is obtain a linear equation that relates x to y along the line that passes through the two points of intersection: 

<pre><code>−2x(x1center−x2center)−2y(y1center−y2center) = (r1)^2−(r2)^2−((x1center)^2−(x2center)^2)−((y1center)^2−(y2center)^2).
</code></pre>

From this you obtain a formula for y in terms of x, then substitute y into one of your circle formulas to obtain a quadratic for x. 
If you don't want to implement a quadratic equation solver, you can use <a href="https://docs.scipy.org/doc/numpy/reference/generated/numpy.roots.html#numpy-roots" rel="nofollow noreferrer">numpy.roots</a> like so:

<pre class="lang-py prettyprint-override"><code>root_array = np.roots(quadratic_coeff, linear_coeff, constant_coef)
</code></pre>

blocks|key|3955422|text|数学/几何问题的语言错误。这是它在更合适的语言(WL)中的样子|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3955423|+Circle+@@@+Thread+@+{RandomReal[{-1,1},{3,2}],+RandomReal[{.5,1},3]}+//+
+Graphics[{
++++#,+Red,+
++++RegionIntersection+@@@+#~Subsets~{2}
+}]&|code-block|syntax|javascript|3955424|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Wrong language for math/geometry things. This is what it looks like in a more appropriate language (WL)

<pre><code> Circle @@@ Thread @ {RandomReal[{-1,1},{3,2}], RandomReal[{.5,1},3]} // 
 Graphics[{
 #, Red, 
 RegionIntersection @@@ #~Subsets~{2}
 }]&amp;
</code></pre>

blocks|key|3955462|text|上面绘制交点的代码没有正确地绘制交点。我调整了代码以绘制它们，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3955463|import+matplotlib.pyplot+as+plt
import+math

def+get_intersections(x0,+y0,+r0,+x1,+y1,+r1):
++++#+circle+1:+(x0,+y0),+radius+r0
++++#+circle+2:+(x1,+y1),+radius+r1

++++d=math.sqrt((x1-x0)**2+%2B+(y1-y0)**2)

++++#+non+intersecting
++++if+d+>+r0+%2B+r1+:
++++++++return+{}
++++#+One+circle+within+other
++++if+d+<+abs(r0-r1):
++++++++return+{}
++++#+coincident+circles
++++if+d+==+0+and+r0+==+r1:
++++++++return+{}
++++else:
++++++++a=(r0**2-r1**2%2Bd**2)/(2*d)
++++++++h=math.sqrt(r0**2-a**2)
++++++++x2=x0%2Ba*(x1-x0)/d+++
++++++++y2=y0%2Ba*(y1-y0)/d+++
++++++++x3=x2%2Bh*(y1-y0)/d+++++
++++++++y3=y2-h*(x1-x0)/d+
++++++++x4=x2-h*(y1-y0)/d
++++++++y4=y2%2Bh*(x1-x0)/d
++++++++return+x3,+y3,+x4,+y4

#+intersection+circles
x0,+y0+=+0,+0
r0+=+5
x1,+y1+=+2,+2
r1+=+5

#+intersecting+with+(x1,+y1)+but+not+with+(x0,+y0)
x2,+y2+=+-1,0
r2+=+2.5

circle1+=+plt.Circle((x0,+y0),+r0,+color='b',+fill=False)
circle2+=+plt.Circle((x1,+y1),+r1,+color='b',+fill=False)
circle3+=+plt.Circle((x2,+y2),+r2,+color='b',+fill=False)

fig+=+plt.figure(figsize+=+(10,+10))+
plt.grid(True)
ax+=+fig.add_subplot(111)

ax.set_xlim((-10,+10))
ax.set_ylim((-10,+10))
ax.add_artist(circle1)
ax.add_artist(circle2)
ax.add_artist(circle3)

intersections1+=+get_intersections(x0,+y0,+r0,+x1,+y1,+r1)
print(intersections1)
if+len(intersections1)+>+0:
++++print(intersections3)
++++i_x3,+i_y3,+i_x4,+i_y4+=+intersections1+
++++#plt.plot([i_x3,+i_x4],+[i_y3,+i_y4],+'o',+color='r')
++++ax.scatter([i_x3,+i_x4],[i_y3,+i_y4]+,marker+='X',s=300,alpha=1)

intersections2+=+get_intersections(x0,+y0,+r0,+x2,+y2,+r2)
print(intersections2)
if+len(intersections2)+>+0:
++++i_x3,+i_y3,+i_x4,+i_y4+=+intersections2+
++++plt.plot([i_x3,+i_x4],+[i_y3,+i_y4],+'o',+color='r')
++++ax.scatter([i_x3,+i_x4],[i_y3,+i_y4]+,marker+='X',s=300,alpha=1)

intersections3+=+get_intersections(x1,+y1,+r1,+x2,+y2,+r2)

if+len(intersections3)+>+0:
++++print(intersections3)
++++i_x3,+i_y3,+i_x4,+i_y4+=+intersections3+
++++#plt.plot([i_x3,+i_x4],+[i_y3,+i_y4],+'o',+color='r')
++++ax.scatter([i_x3,+i_x4],[i_y3,+i_y4]+,marker+='X',s=300,alpha=1)
plt.gca().set_aspect('equal',+adjustable='box')|code-block|syntax|javascript|3955464|输出如图所示：|3955465|​|3955466|📷|atomic|offset|length|3955467|3955468|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/82c798b4b08e97b73090e4fa7f744e98.png|imageAlt^0|0|0|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|11|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|12|8|@]|9|@]|A|$]]|$1|K|3|L|5|M|7|13|8|@]|9|@$N|14|O|15|1|16]]|A|$]]|$1|P|3|J|5|6|7|17|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X|Y|-4]]]]

The above code of plotting the intersection points did not plot the intersecting points properly.
I have adjusted the code to plot them as follows:
<pre><code>import matplotlib.pyplot as plt
import math

def get_intersections(x0, y0, r0, x1, y1, r1):
 # circle 1: (x0, y0), radius r0
 # circle 2: (x1, y1), radius r1

 d=math.sqrt((x1-x0)**2 + (y1-y0)**2)

 # non intersecting
 if d &gt; r0 + r1 :
 return {}
 # One circle within other
 if d &lt; abs(r0-r1):
 return {}
 # coincident circles
 if d == 0 and r0 == r1:
 return {}
 else:
 a=(r0**2-r1**2+d**2)/(2*d)
 h=math.sqrt(r0**2-a**2)
 x2=x0+a*(x1-x0)/d 
 y2=y0+a*(y1-y0)/d 
 x3=x2+h*(y1-y0)/d 
 y3=y2-h*(x1-x0)/d 
 x4=x2-h*(y1-y0)/d
 y4=y2+h*(x1-x0)/d
 return x3, y3, x4, y4

# intersection circles
x0, y0 = 0, 0
r0 = 5
x1, y1 = 2, 2
r1 = 5

# intersecting with (x1, y1) but not with (x0, y0)
x2, y2 = -1,0
r2 = 2.5

circle1 = plt.Circle((x0, y0), r0, color='b', fill=False)
circle2 = plt.Circle((x1, y1), r1, color='b', fill=False)
circle3 = plt.Circle((x2, y2), r2, color='b', fill=False)

fig = plt.figure(figsize = (10, 10)) 
plt.grid(True)
ax = fig.add_subplot(111)

ax.set_xlim((-10, 10))
ax.set_ylim((-10, 10))
ax.add_artist(circle1)
ax.add_artist(circle2)
ax.add_artist(circle3)

intersections1 = get_intersections(x0, y0, r0, x1, y1, r1)
print(intersections1)
if len(intersections1) &gt; 0:
 print(intersections3)
 i_x3, i_y3, i_x4, i_y4 = intersections1 
 #plt.plot([i_x3, i_x4], [i_y3, i_y4], 'o', color='r')
 ax.scatter([i_x3, i_x4],[i_y3, i_y4] ,marker ='X',s=300,alpha=1)

intersections2 = get_intersections(x0, y0, r0, x2, y2, r2)
print(intersections2)
if len(intersections2) &gt; 0:
 i_x3, i_y3, i_x4, i_y4 = intersections2 
 plt.plot([i_x3, i_x4], [i_y3, i_y4], 'o', color='r')
 ax.scatter([i_x3, i_x4],[i_y3, i_y4] ,marker ='X',s=300,alpha=1)

intersections3 = get_intersections(x1, y1, r1, x2, y2, r2)

if len(intersections3) &gt; 0:
 print(intersections3)
 i_x3, i_y3, i_x4, i_y4 = intersections3 
 #plt.plot([i_x3, i_x4], [i_y3, i_y4], 'o', color='r')
 ax.scatter([i_x3, i_x4],[i_y3, i_y4] ,marker ='X',s=300,alpha=1)
plt.gca().set_aspect('equal', adjustable='box')
</code></pre>
The output is given in the image:
<a href="https://i.stack.imgur.com/hVC1R.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hVC1R.png" alt="https://i.stack.imgur.com/KAYeC.png" /></a>

I'm trying to find the intersections between two circles in Python(using Matplotlib) but can't get any values back.

I'm doing this by creating lists of X's and Y's for each individual circle(Matplotlib takes the first argument as X values and the second one as Y values when drawing a circle), and then intersecting the lists accordingly(e.g., circle1 x values with circle2 x values).

<pre><code>import numpy
import math
import matplotlib.pyplot as plt
import random

def origin_circle():
 global x_points
 global y_points
 global r
 global n
 r=1
 n=2**16
 x_points=[(r*math.cos(t)) for t in numpy.linspace(0, 2*numpy.pi*r, n+1)]
 y_points=[(r*math.sin(t)) for t in numpy.linspace(0, 2*numpy.pi*r, n+1)]

def new_circle(x_offset, y_offset):
 global x_points1
 global y_points1
 x_points1=[x_offset+(r*math.cos(t)) for t in numpy.linspace(0, 2*numpy.pi*r, n+1)]
 y_points1=[y_offset+(r*math.sin(t)) for t in numpy.linspace(0, 2*numpy.pi*r, n+1)]

origin_circle()
new_center= random.randint(0, len(x_points))
x_offset = x_points[new_center]
y_offset = y_points[new_center]
new_circle(x_offset, y_offset)
print(set(x_points1).intersection(set(x_points)))
print(set(y_points1).intersection(set(y_points)))
</code></pre>

I expected to get values back, but the set that returned was empty.

Finding the intersection of two circles

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我试图在Python中找到两个圆之间的交点(使用Matplotlib)，但无法获得任何值。为此，我为每个单独的圆创建了X和Y的列表(当绘制圆时，Matplotlib将第一个参数作为X值，第二个参数作为Y值)，然后相应地将列表相交(例如，circle1 x值与circle2 x值)。import numpyimport ...

问寻找两个圆的交点
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