The common length of the two circles obtained by + 1y + 10Y + 2x = 0-2

The common length of the two circles obtained by + 1y + 10Y + 2x = 0-2

Subtracting two formulas
16x+12y-40=0
4x-3y-10=0
This is the line where the common string is
Y = (4x-10) / 3 is substituted into x ^ 2 + y ^ 2-10x-10y = 0 and sorted out
5x^2-58x+80=0
x1+x2=58/5,x1x2=16
(x1-x2)^2=(x1+x2)^2-4x1x2=1764/25
y1-y2=[(4x1-10)/3]-[(4x2-10)/3]=4(x1-x2)/3
(y1-y2)^2=16(x1-x2)^2/9=3136/25
So chord length = √ [(x1-x2) ^ 2 + (y1-y2) ^ 2] = 14

It is known that two circles x2 + y2-10x-10y = 0, X2 + Y2 + 6x-2y-40 = 0, Find (1) the equation of the line where their common chord lies; (2) the common chord length

（1）x2+y2-10x-10y=0，①；x2+y2+6x-2y-40=0②；
② - 1: 2x + Y-5 = 0 is the equation of the line where the common chord is located;
(2) The chord center distance is: | 10 + 5 − 5|
22+12=
20, half of the chord length is
50−20＝
30, the common chord length is 2
Thirty

Find the common chord length of the circle x 2 + y 2 - 10 x - 10 y = 0 and x 2 + y 2 - 6 x + 2 y - 40 = 0 Please give the complete process

The two equations about a circle are arranged into the simplest form by matching method: (X-5) 2 + (Y-5) 2 = 50; (1) (x-3) 2 + (y + 1) 2 = 50; (2) two groups of values of X and y are solved, and then the chord length can be obtained by using the distance formula of two points in analytic geometry
This problem can also directly find the common solution of the above equation

Find the position relation of circle x2 + y2-10x-10y = 0 and circle x2 + y2-6x + 2y-40 = 0

(x-5)^2+(y-5)^2=50
(x-3)^2+(y+1)^2=50
Radius is root 50
Center (5,5) and (3, - 1)
Center distance (5-3) ^ 2 + (5 + 1) ^ 2
That is, the root 40 is less than the radius and the intersection

Find the common chord length of circle x ^ 2 + y ^ 2-10x-10y = 0 and circle x ^ 2 + y ^ 2-6x + 2y-40 = 0

The equation of two upper circles is converted into standard form, and the following results are obtained: (X-5) ^ 2 + (Y-5) ^ 2 = 50, (x-3) ^ 2 + (y + 1) ^ 2 = 50, so the centers of two circles are (5,5), (3, - 1), and the radii of both circles are 5 √ 2. Therefore, the center of two circles and the intersection point of two circles form a rhombus, and the two pairs of angular lines are common chord and center distance

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 available: 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 common chord length of circle x2 + y2-10x-10y0 and circle x2 + y2-6x + 2y-400 It's 2y-40 = 0

By subtracting the equations of two circles, x + 3y-10 = 0 is the line equation of the common chord
Then, by combining the equations of one of the circles, the intersection point a (- 2,4) B (10,0) is obtained
Therefore, the common string | ab | = 4 √ 10

1. Find the circle equation that crosses the intersection point of C1: x ^ 2 + y ^ 2 + 6x-4 = 0 and C2: x ^ 2 + y ^ 2 + 6y-28 = 0 and the center of the circle is in the straight line X-Y-4 = 0 And (y) ^ 2 = ^ 2 3. Find the circle equation with the common chord of C1: x ^ 2 + y ^ 2 + 4x + 1 = 0 and C2: x ^ 2 + y ^ 2 + 2x + 2Y + 1 = 0

First of all, the equations of two circles are connected and subtracted to obtain X-Y + 4 = 0. Then, x = - 1, y = 3 or x = - 6, y = - 2 are obtained. The two points are connected and the center of the circle is on the middle perpendicular line

Find the equation of the circle passing through the intersection of two circles x ^ 2 + y ^ 2 + 6x-4 = 0 and x ^ 2-4x + y ^ 2 = 0 and tangent to the line x-root 3y-6 = 0

If the center of the circle is (a, b), the equation is as follows (- 1-A) (- 1-A) + (3-B) (3-B) = - 6 (- 6-A) (- 6-A) (- 6-A) (- 6-A) (- 2-B) (3-B) = (- 6-A) (- 6-A) + (- 2-B) (- 2-B) a-b-4 = 0 and a = 1 /..., respectively

In the straight line passing through point (2,1), the chord cut by circle x square + y square - 2x + 4Y = 0 is the longest linear equation Please explain why

X square + y square - 2x + 4Y = 0
(x-1)^2+(y+2)^2=5
The C coordinates of the center of the circle are (1, - 2)
Among the lines passing through point a (2,1), the chord cut by circle C is the longest line, which is perpendicular to AC
AC slope = (- 2-1) / (1-2) = 3
Therefore, the slope of the line = - 1 / 3
The linear equation: Y-1 = - 1 / 3 * (X-2)
That is: x + 3y-5 = 0