Problems of circular tangent planes, example problems of circular slices

The answer is 14 parts. See**.

Abstract problem: How many regions can n circles divide a plane into at most.

The general formula is: b(n)=n(n-1)+2

The proof is as follows: 1) How many segments can n circles divide into a circle. The general formula is denoted as a(n).

2) n circles can divide a plane into multiple regions at most. The general term formula is denoted as b(n) the first problem, because two circles intersect at most 2 intersection points, so n circles leave a maximum of 2n intersection points on the specified circle, and divide this circle into 2n segments, i.e., a(n)=2n

The second problem is that if there are already n circles on the plane, and they divide the plane into the most regions, then when the n+1 circle goes down, in order to ensure that the most regions are obtained, this circle is required to intersect with the previous n circles, and the newly generated intersection points do not coincide with the previous intersection points. According to the first problem, the previous n circles divide the n+1 circle into a(n) segments, and each segment divides the original area in two, so b(n+1)=b(n)+a(n), and b(1)=2, a(n)=2n is brought in, and it is easy to get b(n)=n(n-1)+2

You draw three circles to intersect first, and then make up the fourth circle, be careful not to let the lines of the circles cross the intersection point, and then calculate that there should be 14 I don't know if it's right. FYI.

There is a part outside the 3 circles, there are three parts of a single circle, three parts of two circles intersect, and one of the three circles!! 1+3+3+1=8！！！Remember, it's a plane, not a space!!

It's only level 1 and can't get on the picture.。。。

That is, each circle intersects with the other 3 circles, which is the most.

The three centers of the circle intersect in a regular triangle, and the last one intersects the others in the center of the triangle.

It should be the most common situation.

Divide the plane into 14 parts.

1. The circle tangent to all sides of the polygon is called the inscribed circle of the polygon.

2. The circle that intersects all the corners of the polygon is called the circumscribed circle of the polygon.

Example 3 The distance between a point p and a point on o in the plane is a minimum of 3 cm and a maximum of 8 cm, then the radius of the circle is cm.

Example 4 In a circle with a radius of 5 cm, the chord ab cd, ab=6cm, cd=8cm, what is the distance between ab and cd?

Example 6Known: The radius of o is 0a=1, the length of the chord ab and ac is 2,3 respectively, find the degree of bac

Test Center Quick Practice] 1Of the following propositions, the correct one is ( )a three points determine a circle.

b Any triangle has one and only one circumscribed circle.

c Any quadrilateral has an inscribed circle d The outer center of an isosceles triangle must be on its outer 2 If the outer center of a triangle is on one side of it, then this triangle must be ( ) a isosceles triangle b right triangle c equilateral triangle d obtuse triangle.

3 The number of circumscribed triangles of a circle is ( ) a 1 b 2 c 3 d infinite 4 The number of circumscribed triangles of a triangle is ( ) a 1 b 2 c 3 d infinite 5 In the following statement, the correct number is ( ).

Any point can determine a circle; Any two points can determine a circle; Any three points can determine a circle; It can be circled through any point; After any two points, there must be circles A, 1, B, 2, C, 3.

d 4 pcs. 6.The graph composed of points whose distance from the center of the circle is no greater than the radius

a.the outer part of the circle (including the boundary); b.the interior of the circle (excluding the boundary); c.

Round; d.The interior of the circle (including the boundary) 7It is known that the radius of O is 6cm, and P is the midpoint of the line segment Oa, and if the point P is on O, then the length of Oa ( )A

Equal to 6cm bequal to 12cm; c.Less than 6 cm d

More than 12cm

8.As shown in the figure, the diameter of o is 10cm, the chord ab is 8cm, p is a point on the chord ab, and if the length of op is an integer, then the point p that satisfies the condition has ( ) 9As shown in the figure, a is a point within the radius of 5 o, and oa = 3, and the string that crosses the point a and is less than 8 in length has ( ) strips.

Make a common tangent line, then the tangent angle of the pair of chords at the top is equal to the circumferential angle of the same arc, and the internal error angle can be equal, that is, there is a quadrilateral ...... parallel to the opposite side

Probably that's the proof of it, and I can't think of any other way to do it for the time being.