If two circles x2 + y2-10x-10y = 0 and X2 + y2-6x + 2y-40 = 0 intersect at two points, the equation of the line where their common chord is located is______ .

∵ the two circles are x2 + y2-10x-10y = 0 ①, X2 + y2-6x + 2y-40 = 0 ②
② - ① it can be concluded that 4x + 12y-40 = 0
That is, x + 3y-10 = 0
The equation of the line where the common chord of two circles is located is x + 3y-10 = 0
So the answer is: x + 3y-10 = 0

Find the equation of the circle passing through the intersection of two circles x2 + Y2 + 6x-4 = 0 and X2 + Y2 + 6y-28 = 0, and the center of the circle is on the straight line X-Y-4 = 0

The circular equation can be set as a (x ^ 2 + y ^ 2 + 6x-4) + x ^ 2 + y ^ 2 + 6y-28 = 0
Therefore, the center of the circle is at (3a / (a + 1), 3 / (a + 1))
Take him into the linear equation and find a = - 7
Is the equation ^ 7Y + 2 x = 0

Find the equation of the circle passing through the intersection of two circles x ^ 2 + y ^ 2 + 6x-4 = 0 and x ^ 2 + y ^ + 6y-28 = 0 and the center of the circle is on the straight line X-Y-4 = 0

x^2+y^2+6x-4=0
x^2+Y2+6y-28=0
By subtracting the two formulas, we get 6x-6y + 24 = 0 and y = x + 4
Substituting y = x + 4 into any of the above formulas, the coordinates of the two intersections can be obtained as a (- 1,3) B (- 6, - 2)
The slope of AB line is K=-
Then the intersection point of the vertical bisector of line AB and the straight line X-Y-4 = 0 is the center of the circle
The slope of AB vertical bisector is - 1
The coordinates of the midpoint of line AB are (- 7 / 2,1 / 2)
So we can get the straight line equation of the vertical bisector of line AB as y = - x-3
Change y = - x-3
X-Y-4 = 0 to solve the equations, the coordinates of the center of the circle are (1 / 2, - 7 / 2)
Radius of circle = √ [(- 7 / 2-3) ^ 2 + (1 / 2 + 1) ^ 2] = √ 178 / 2
So the equation of the circle is (x-1 / 2) ^ 2 + (y + 7 / 2) ^ 2 = 178 / 4

Find the equation of the circle passing through the intersection of two circles C1: x ^ 2 + y ^ 2 + 6x-4 = 0 and C2: x ^ 2 + y ^ 2 + 6y-28 = 0, and the center of the circle is on the straight line X-Y-4 = 0 Using the equation of common string system of two circles, we can get the linear equation of common string with C1-C2. How to find the radius after solving this

Then find out the distance between the center of the circle and the center of the circle

The linear equation of the common chord of the circle x2 + Y2 + 2x = 0 and X2 + y2-4y = 0 is______ .

Circle x2 + Y2 + 2x = 0 ① And X2 + y2-4y = 0 II.
① (2) x + 2Y = 0 is the linear equation of the common chords of the circles x2 + Y2 + 2x = 0 and X2 + y2-4y = 0
So the answer is: x + 2Y = 0

The common chord length of circle x2 + y2-4 = 0 and circle x2 + y2-4x + 4y-12 = 0 is () A. Two B. Three C. 2 Two D. 3 Two

By subtracting the equation of circle x2 + y2-4 = 0 and circle x2 + y2-4x + 4y-12 = 0, X-Y + 2 = 0,
∵ the distance from the center of the circle (0, 0) to the straight line X-Y + 2 = 0 d = 2
2=
2，r=2，
Then the common chord length is 2
r2−d2=2
2．
Therefore, C

Find the common chord length of the square of circle x + the square of Y - 4 = 0 and the square of circle x + the square of Y - 4x + 4y-12 = 0

By simplifying the circle equation, we know that the center of circle 1 is (0,0), and the radius is 2
Circle 2 has a center of (2,2) and a radius of 2
Distance between two centers = √ [2 ^ 2 + 2 ^ 2] = 2 √ 2
Therefore, in a circle, the distance from the center of the circle to the common chord = 2 √ 2 / 2 = √ 2
From Pythagorean theorem
Semi common chord length = √ [2 ^ 2 - (√ 2) ^ 2] = √ 2
So the common chord length = 2 √ 2

If the common chord length of two circles x ^ 2 + y ^ 2 + 4x-4y = 0 and x ^ 2 + y ^ 2 + 2x-12 = 0 is

The center of the circle x ^ 2 + y ^ 2 + 2x-12 = 0 is (- 1,0), and the radius is √ 13,
The equation of intersecting chords of two circles is to subtract the equations of two circles to get x-2y + 6 = 0;
The distance from (- 1,0) to x-2y + 6 = 0 is d = 5 / √ 5 = √ 5
According to Pythagorean theorem, the chord length is 2 * √ (13-5) = 4 √ 2

The midpoint of a chord of ellipse x ^ 2 + 4Y ^ 2 = 16 is (3,1) to find the linear equation of the chord

Let the chord midpoint m (3,1), the chord is ab, a (x1, Y1), B (X2, Y2), (x1 + x2) / 2 = 3, (Y1 + Y2) / 2 = 1, and substituting the coordinates of a and B into the elliptic equation respectively, we get [(y2-y1) / (x2-x1)] * ((Y1 + Y2) / (x1 + x2)), (y2-y1) / (x2-x1) = - 1 / 12, the slope of the straight line k = - 1 / 12, and the straight line passes through (3,1)

In this paper, we find the equation of the line where the chord is located with a point a (1, - 1) in the ellipse x * 2 + 4Y * 2 = 16 as the midpoint Not very good at it I hope I can teach you Can we use the polar coordinates to solve the problem?

Let the line y + 1 = K (x-1) be a (x1, Y1) B (X2, Y2) at the intersection of the circle
The equation about X is obtained by connecting two equations, and then X1 + X2 is obtained by using Veda theorem=
These relations... Then X1 + x2 = 2, Y1 + y2 = - 2, these relations can be solved
The process is too complicated``

If two circles x2 + y2-10x-10y = 0 and X2 + y2-6x + 2y-40 = 0 intersect at two points, the equation of the line where their common chord is located is______ .