Given the circle C1: x square + y square + 2x + 6y + 9 = 0 and circle C2: x square + y square - 6x + 2Y + 1 = 0, find the common tangent equation of circle C1 and circle C2

Given the circle C1: x square + y square + 2x + 6y + 9 = 0 and circle C2: x square + y square - 6x + 2Y + 1 = 0, find the common tangent equation of circle C1 and circle C2

C1: (x + 1) ^ 2 + (y + 3) ^ 2 = 1 and circle C2: (x-3) ^ 2 + (y + 1) ^ 2 = 9
Let the common tangent y = KX + B, and change it into kx-y + B = 0, then the distance from the center of two circles to the common tangent is 1 and 3 respectively. From the formula of distance from point to line, we get ABS (- K + 3 + b) / SQR (k ^ 2 + 1) = 1 and ABS (3K + 1 + b) / SQR (k ^ 2 + 1) = 3
The solution is: k = 0, B = - 4 or K = 4 / 3, B = 0 or K = - 3 / 4, B = - 5 / 2;
The tangent is y = - 4, y = 4 / 3x, y = - 3 / 4x-5 / 2, and the y-axis is also a common tangent

Given the circle C1: x ^ 2 + y ^ 2 = 1, circle C2: X62 + y ^ 2-2x-2y + 1 = 0, try to find the equation of the line where the common chord of two circles is located

Subtract the former from the latter. X + y = 1,

It is known that the circle C1: x ^ 2 + y ^ 2 + 2x + 2y-8 = 0 and circle C2: x ^ 2 + y ^ 2-2x + 10y-24 = 0 intersect at two points a and B And the length of ab

The common string equation x ^ 2 + y ^ 2 + 2x + 2y-8 - (x ^ 2 + y ^ 2-2x + 10y-24) = 0, that is, x-2y + 4 = 0. The standard formula of circle C1 is (x + 1) ^ 2 + (y + 1) ^ 2 = 10, so the center of the circle is (- 1, - 1), and the radius is r = √ 10, so the distance from the center of the circle to the common chord is d = | - 1 + 2 + 4 | / √ (1 ^ 2 + 2 ^ 2) = √ 5, so AB / 2 = √ (10-5) =

If the circle C1: x2 + y2-10x-10y = 0 and C2: x2 + Y2 + 6x + 2y-40 = 0 intersect at a and B, then the length of common chord AB is () A. 5 B. 5 Two C. 5 Three D. 10

The equation of common string obtained by subtracting two circles is 4x + 3y-10 = 0
∵ x2 + y2-10x-10y = 0, the center coordinates are (5, 5), and the radius is 5
Two
The distance from the center of the circle to the common chord is d = | 20 + 15 − 10|
5=5
∴AB=2
(5
2)2−52=10
Therefore, D

Given the circle C1: x ^ 2 + y ^ 2-3x-3y + 3 = 0, circle C2: x ^ 2 + y ^ 2-2x-2y = 0, find the line equation and chord length of the common chord of two circles

A conclusion is used: the equation of common string is the subtraction of two circle equations. X ^ 2 + y ^ 2-3x-3y + 3 - (x ^ 2 + y ^ 2-2x-2y) = 0 = > - X-Y + 3 = 0, that is, x + Y-3 = 0 This is the equation of common string, circle C2 (x-1) ^ 2 + (Y-1) ^ 2 = 2, r = radical 2, Center (1,1), distance from center (1,1) to straight line: D = | 1 + 1-3 | / √ (1 ^ 2 +...)

The linear equation of the common chord of the circle C1: x ^ 2 + y ^ 2-2x + 10y-20 = 0 and C2: x ^ 2 + y ^ 2 + 2x + 2y-8 = 0 is

x²+y²-2x+10y-20=0
x²+y²+2x+2y-8=0
2 minus
4x-8y+12=0
x-2y+3=0
That's what you want

The line where the common chord of circle C1: x2 + y2 = 1 and circle C2: x2 + y2-2x-2y + 1 = 0 is surrounded by circle C3: (x-1) 2 + (Y-1) 2 = 25 The length of the chord cut is 4___ .

The linear equation of the common chord of circle C1 and circle C2 is: x2 + y2-1 - (x2 + y2-2x-2y + 1) = 0, that is, x + Y-1 = 0,
The distance between the center of circle C3 (1,1) and the straight line x + Y-1 = 0 d = | 1 + 1-1|
2=
Two
2，
So the chord length is 2
r2-d2=2
Twenty-five
4-1
2=
23，
So the answer is
23．

What is the line equation of the common chord of two circles x ^ 2 + y ^ 2-2x-3 = 0 and x ^ 2 + y ^ 2 + 6y-1 = 0? Is this inference? No restrictions on use?

Common line of string equation
The simplest solution is to make difference between two equations
6y+2x+2=0

The linear equation of chord length 8 passing through point a (- 1,10) and cut by circle x2 + y2-4x-2y-20 = 0 is______ .

The circle x2 + y2-4x-2y-20 = 0 is transformed into the standard equation (X-2) 2 + (Y-1) 2 = 25
When the slope of the line is k, the equation of the line is Y-10 = K (x + 1), that is, kx-y + K + 10 = 0
The distance from the center of a circle (2, 1) to the straight line d = | 2K − 1 + K + 10|
k2+1＝|3k+9|
k2+1
And ∵ chord length is 8, circle radius r = 5, ᙽ chord center distance d = 3,
∴|3k+9|
k2+1＝3，
∴k＝−4
Three
The linear equation is 4x + 3y-26 = 0
When the slope of the line does not exist, the equation is x + 1 = 0, and the distance from the center of the circle (2,1) to the line is 3, and the chord length is 8
To sum up, the equation of the straight line is 4x + 3y-26 = 0 or x = - 1
So the answer is: 4x + 3y-26 = 0 or x = - 1

The linear equation of chord length 8 passing through point a (- 1,10) and cut by circle x2 + y2-4x-2y-20 = 0 is______ .

The circle x2 + y2-4x-2y-20 = 0 is transformed into the standard equation (X-2) 2 + (Y-1) 2 = 25
When the slope of the line is k, the equation of the line is Y-10 = K (x + 1), that is, kx-y + K + 10 = 0
The distance from the center of a circle (2, 1) to the straight line d = | 2K − 1 + K + 10|
k2+1＝|3k+9|
k2+1
And ∵ chord length is 8, circle radius r = 5, ᙽ chord center distance d = 3,
∴|3k+9|
k2+1＝3，
∴k＝−4
Three
The linear equation is 4x + 3y-26 = 0
When the slope of the line does not exist, the equation is x + 1 = 0, and the distance from the center of the circle (2,1) to the line is 3, and the chord length is 8
To sum up, the equation of the straight line is 4x + 3y-26 = 0 or x = - 1
So the answer is: 4x + 3y-26 = 0 or x = - 1