The chord length of a line passing through the origin and with an inclination angle of 60 ° cut by the circle x2 + y2-4y = 0 is () A. Three B. 2 C. Six D. 2 Three

The equation of circle x2 + y2-4y = 0 can be transformed into:
x2+（y-2）2=4，
That is, the center of the circle is a (0, 2), and the radius is r = 2,
The distance from a to the straight line on, that is, the chord center distance is 1,
∴ON=
3，
The chord length is 2
3，
Therefore, D

The chord length of a line passing through the origin and with an inclination angle of 60 ° cut by the circle x2 + y2-4y = 0 is () A. Three B. 2 C. Six D. 2 Three

The equation of a circle that passes through the intersection point of the straight line x-2y + 4 = 0 and the circle x square + y square + 2x-4y + 1 = 0 and satisfies the following conditions: 1) the point passing through the circle 2) has the minimum area

This problem does not have any special thinking characteristics, so we should follow the normal way of thinking. First, we put the straight line into the circle to solve the coordinates of the two intersecting points. Then, according to the conditions, once the origin is crossed, the circle is determined according to three points, and the three parameters of ABC are solved according to the standard equation of the circle, In this way, it will be very troublesome to solve ABC. The result I made doesn't seem to be the result of a scientific topic. So I doubt that either your topic is written or the topic is wrong, so I don't suggest you have to make the result and understand the solution
The minimum area in the second question is actually the minimum of C (that is, the minimum radius of the circle). After substituting the coordinates of the intersection point into the set equation of the circle, an equation about C can be solved. According to the condition (generally the most basic theoretical condition in quadratic function), because the square of C is on the right side of the equal sign in the standard formula of the circle, the square root must be greater than or equal to 0, This has the limit condition, solves a and B successively, should be able to obtain the result
If this kind of question is not a model test or textbook, there is no need to make the results. One is a waste of time, the other is not to learn any new ideas and methods. It is a waste of time
I don't know if there is any expert who can make the result. Anyway, the result I made is not an integer, which is not "simple"

Find the equation of the circle that passes through the intersection point of the straight line x-2y + 4 = 0 and the circle x ^ 2 + y ^ 2 + 2x-4y + 1 = 0, and passes through the origin

The method of this problem is very clever, using the circle system
Crossing the intersection of a line and a circle
The circle can be expressed as x ^ 2 + y ^ 2 + 2x-4y + 1 + K (x-2y + 4) = 0
(it is shown that the formula is quadratic and must be a circle. When x and Y take the coordinates of the intersection points, the formula is bound to hold, so the circle represented by the formula must pass through two intersection points.)
Bring in the coordinates of the circle (0,0)
1+4k=0
k=-1/4
Replace the original form
The circle is
x^2+y^2+2x-4y+1-0.25x+0.5y-1=0
That is, x ^ 2 + y ^ 2 + 1.75x-3.5y = 0
To standard equation
(x+7/8)^+(y-7/4)^=245/64

Find the equation of the circle where the line x-2y + 4 = 0 and the circle x ^ 2 + y ^ 2 + 2x-4y + 1 = 0 and satisfy one of the following conditions: 1. There is a minimum area through the origin 2

The radius of a circle with o as the center is x ^ 2 + y ^ 2 = 20. Therefore, we can solve the equations as follows: x ^ 2 + y ^ 2 + 8x-4y = 0 and x ^ 2 + y ^ 2 = 20

The equation of a circle passing through the intersection of the straight line 2x + y + 4 = 0 and the circle x2 + Y2 + 2x-4y + 1 = 0 and satisfying one of the following conditions: (1) Crossing the origin (2) There is a minimum area

The equation of the circle passing through the intersection of the straight line 2x + y + 4 = 0 and the circle x2 + Y2 + 2x-4y + 1 = 0 can be set as (x2 + Y2 + 2x-4y + 1) + λ (2x + y + 4) = 0
(1) Substituting (0, 0) into, we can get 1 + 4 λ = 0, νλ = - 1
4，
The equation of circle is x2 + Y2 + 3
2x−17
4y＝0；
(2) (x2 + Y2 + 2x-4y + 1) + λ (2x + y + 4) = 0 can be transformed into x2 + Y2 + (2 λ + 2) x + (λ - 4) y + 1 + 4 λ = 0
The radius of the circle is
(2λ+2)2+(λ−4)2−4(1+4λ)
4=
Five
4(λ−8
5)2+4
Five
∴λ=8
At 5, the radius is the smallest and the area is the smallest,
So the equation of the circle is (x + 13)
5)2+(y−6
5)2＝4
Five

Find the circle that crosses the intersection point of the straight line 2x + y + 4 = 0 and the circle x ^ 2 + y ^ 2 + 2x-4y + 1 = 0 and passes through the origin

Let x ^ 2 + y ^ 2 + 2x-4y + 1 + λ (2x + y + 4) = 0
Substituting the origin (0,0) into 1 + 4 λ = 0
So λ = - 1 / 4
So the circle is x ^ 2 + y ^ 2 + 2x-4y + 1 + (- 1 / 4) * (2x + y + 4) = 0
If it is standardized: (x + 3 / 4) ^ 2 + (y-17 / 8) ^ 2 = 325 / 64
[supplementary knowledge]
Through the intersection point of the straight line ax + by + C = 0 and the circle x ^ 2 + y ^ 2 + DX + ey + F = 0, the circular system equation is: x ^ 2 + y ^ 2 + DX + ey + F + λ (AX + by + C) = 0
If you do not understand, please hi me, I wish you a happy study!

The equation of a straight line with chord length of root 3 obtained from the circle x ^ 2 + y ^ 2-2x = 0 is

The square of the distance from the center of the circle to the string is 1 ^ 2 - [(root 3) / 2] ^ 2 = 1-3 / 4 = 1 / 4. That is, the distance from the center of the circle to the string is 1 / 2

It is known that the circle x side +y side =25, O is the coordinate origin, the chord length of the line L passing through the point P (0,3 times of the root sign 2) cut by the circle is 8, and the equation of the line L is obtained

Let the slope of the straight line l be K, then the equation of L is y = KX + 3 √ 2, that is, kx-y + 3 √ 2 = 0
Let the chord of the line L cut by ⊙ o be AB, and then let the midpoint of AB be c. obviously, there are: OC ⊥ AC, OA = 5, AC = 4,
According to the Pythagorean theorem, OC = √ (OA ^ 2-ac ^ 2) = √ (25-16) = 3
The formula of the distance from a point to a straight line is as follows: 3 √ 2 / √ (k ^ 2 + 1) = 3, ν√ (k ^ 2 + 1) = √ 2, νk = 1 or K = - 1
There are two equations of the line L satisfying the conditions, which are y = x + 3 √ 2 and y = - x + 3 √ 2

The chord length of a line passing through the origin and with an inclination angle of 60 ° cut by the circle x2 + y2-4y = 0 is () A. Three B. 2 C. Six D. 2 Three