Given that two circles C1: x? + y? = 4, C2: x? + y? - 2x-4y + 4 = 0, and the straight line L: x + 2Y = 0, find the passing circle C1 and C Find the equation of the circle which passes through the intersection of C1 and C2 and tangent to the line L Please don't copy the answer from others on Baidu. I said that I can't understand why he set the equation of circle like that. Our teacher has never taught us to set it like that

Given that two circles C1: x? + y? = 4, C2: x? + y? - 2x-4y + 4 = 0, and the straight line L: x + 2Y = 0, find the passing circle C1 and C Find the equation of the circle which passes through the intersection of C1 and C2 and tangent to the line L Please don't copy the answer from others on Baidu. I said that I can't understand why he set the equation of circle like that. Our teacher has never taught us to set it like that

Circle C1: x ^ 2 + y ^ 2 = 4
Circle C2: x ^ 2 + y ^ 2-2x-4y + 4 = 0
The linear equation of common chord is obtained by subtracting two circle equations
2x+4y-4=4
x+2y-4=0
X = 4-2y into C1 circle equation:
16-16y+4y^2+y^2=4
5y^2-16y+12=0
(5y-6)(y-2)=0
y1=6/5,y2=2
The corresponding solutions are as follows
x1=8/5,x2=0
So the intersection of the two circles is:
(8/5,6/5）、(0,2)
Let the center of the circle to be p (x, y)
Then the distance between P and a, B, l is equal, and is the radius r of the circle
R ^ 2 = (x + 2Y) ^ 2 / 5 (distance from L)
R ^ 2 = x ^ 2 + (Y-2) ^ 2 (distance from a)
R ^ 2 = (X-8 / 5) ^ 2 + (y-6 / 5) ^ 2 (distance from b)
The solution is x = 1 / 2 y = 1 R ^ 2 = 5 / 4
Therefore, the circle is (x-1 / 2) ^ 2 + (Y-1) ^ 2 = 5 / 4

The positional relationship between two circles C1: x? + y? = 2 and C2: x? + y? - 2x-1 = 0

Because the center of the circle x? 2 + y? 2 = 2 is the origin and the radius is the root 2
The radius of the circle is (1,0), and the radius is root 2,
The distance between the two centers is 1,1

It is known that the circle C1: x? + y? - 4x-2y-3 = 0, the circle C2: x? + y? - 2x + M = 0, where - 5 < m < 1 ① If M = - 1, judge the position relationship between C1 and C2, and find the common tangent equation of two circles ② Let the line where the common chord of circle C1 and circle C2 are located is l, and the distance between the center of circle C2 and the straight line L is 2 / 2 of the root, then the equation of line L and the common chord length are obtained

① When m = - 1, the center C1 (2,1), radius R1 = 2 √ 2, the center C2 (1,0), radius R2 = √ 2, center distance C1C2 = √ 2 = R1-R2, so the two circles are tangent. Because the two circles are tangent, the common tangent equation of two circles can be obtained by subtracting the equations of two circles: x + y + 1 = 0

It is known that the circle C1: x? + y? + 2x + 2y-8 = 0 and C2: x? + y? - 2x + 10y-24 = 0 intersect at two points a and B Find the equation of the circle whose center is on the straight line y = - X and passes through two points a and B

Curve system solution
Let the circle: C1 + α C2 = 0
The center of the circle is ((2 α - 2) / 2 (1 + α), (10 α - 2) / 2 (1 + α))
And the center of the circle is on y=-x
Therefore, α = 1 / 3
So the circle is

Given that the circle C1: x2 + Y2 + 2x + 2y-8 = 0 and C2: x2 + y2-2x + 10y-24 = 0 intersect at two points a and B, find the length of ab

The equation of AB is 4x-8y + 16 = 0, that is, x-2y + 4 = 0
X2 + Y2 + 2x + 2y-8 = 0, that is (x + 1) ^ 2 + (y + 1) ^ 2 = 10
X2 + y2-2x + 10y-24 = 0, that is (x-1) ^ 2 + (y + 5) ^ 2 = 50
The distance from the center of a circle (- 1, - 1) to the straight line x-2y + 4 = 0, the radius of the circle, half of AB, forms a right triangle
According to Pythagorean theorem
[|-1+2+4|/√5]^2+1/4AB^2=10
AB=6

It is known that the circle C1: x2 + Y2 + 2x + 2y-8 = 0 and C2: x2 + y2-2x + 10y-24 = 0 intersect at two points (1) of a and B to find the length of line ab

The length of AB is 2 √ 5. Calculation method: the equation group is composed of two equations. From C1-C2, x = 2y-4 can be obtained. If this formula is applied to C1 formula, y = 0 or 2 can be obtained, thus x = - 4 or 0. That is, the length of ab can be 2 √ 5 when a and B are (- 4,0) and (0,2)

Given that two circles C1: x ^ 2 + y ^ 2 = 4, C2: x ^ 2 + y ^ 2-2x-4y + 4 = 0, and the straight line L: X-Y-4 = 0, the equation of the circle passing through the intersection of two circles and tangent to the line L is solved

(using the equation of circle system) from the meaning of the title, we can assume that the equation of the circle is (x? 2 + y? - 2x-4y + 4) + T (x? 2 + y? - 4) = 0. (t ≠ - 1). The standard equation is [X-1 / (1 + T)] + [Y-2 / (T + 1)]? = (4T? + 1) / (1 + T) 2

Find the equation of a circle passing through the intersection of circle C1: x2 + y2-4x + 2Y + 1 = 0 and circle C2: x2 + y2-6x = 0 and passing through the point (2, - 2)

Let the equation of the circle passing through the intersection point of circle C1: x2 + y2-4x + 2Y + 1 = 0 and circle C2: x2 + y2-6x = 0 is (x2 + y2-4x + 2Y + 1) + λ (x2 + y2-6x) = 0,
Substituting the point (2, - 2), we can get (4 + 4-8-4 + 1) + λ (4 + 4-12) = 0,
∴λ=-3
4，
The equation of the circle is (x2 + y2-4x + 2Y + 1) - 3
4 (x2 + y2-6x) = 0, that is, X2 + Y2 + 2x + 8y + 4 = 0

Find the equation of the circle with the smallest radius passing through the intersection of two circles: circle C: x2 + y2 = 4x + y + 1 = 0 and circle C2: x2 = Y2 + 2x + 2Y + 1 = 0?

By subtracting the equations of two circles, the straight line equation 2x-y = 0 of the intersection point is obtained. Bring y = 2x into one of the circles to find the intersection point (x1, Y1) (X2, Y2)
Now if you want the circle with the smallest radius, the circle is the circle with the diameter of the intersection point
The center of the circle is (x1 + x2 / 2, Y1 + Y2 / 2). The radius is half the distance of the intersection point

Given the circle C1: x2 + Y2 + 2x + 2y-8 = 0 and C2: x2 + y2-2x + 10y-24 = 0, we can find the equation of the common tangent of two circles

X－2Y+4=0
Subtracting two circles is the common tangent equation