Find the length of the common chord of the circle x2 y2 4 0 and the circle x2 y2 4x 4y 12 0

x2+y2-4=0（1）

x2+y2-4x+4y-12=0（2）

1)-(2) De.

4x-4y+8=0

y-2 (3)

3) Substitution (1) obtained.

y^2-4y+4+y^2-4=0

2y^2-4y=0

The solution yields y1=0

y2=2 is substituted into equation (3).

x1=-2x2=0

So common chord length = [(x1-x2) 2+(y1- y2) 2]= 8=2 2

Let the common chord be ab

x²+y²-4=0 ……

x²+y²-4x+4y-12=0 ……

Get: 4x-4y+8=0 ......

That is: x-y+2=0, which is the straight line where the common chord is. (If you don't understand, see the notes below).

The center of the circle x + y -4 = 0 is o(0,0) and the radius r = 2 is the distance from the center of the circle o to the common chord .

d=2/√2=√2

ab|/2=√(r²-d²)=√2

ab|=2√2

Note: Let the root of the system of equations be a(x1,y1) and b(x2,y2), and a can be substituted into , .

Then use 4x1-4y1+8=0, that is, x1-y1+2=0, and x2-y2+2=0 can be obtained in the same way

a(x1,y1),b(x2,y2) satisfy the equation x-y+2=0ab on the straight line x-y+2=0.

Let the common chord be ab

x²+y²-4=0 ……

x²+y²-4x+4y-12=0 ……

Get: 4x-4y+8=0 ......

That is: x-y+2=0, which is the straight line where the common chord is. (If you don't understand, see the notes below).

The center of the circle x + y -4 = 0 is o(0,0), and the radius r = 2 is the distance from the center of the circle o to the common chord.

d=2/√2=√2

ab|/2=√(r²-d²)=2

ab|=2√2

Note: Let the root of the system of equations be a(x1,y1) and b(x2,y2), and a can be substituted into , .

Then use 4x1-4y1+8=0, that is, x1-y1+2=0, and x2-y2+2=0 can be obtained in the same way

a(x1,y1),b(x2,y2) satisfies the equation x-y+2=0ab on the straight line x-y+2=0, 4,

The center of the circle of x 2+y 2+2x-12=0 is (-1,0), the radius is 13, and the equation of the intersecting chord of the two circles is to subtract the equations of the two circles to obtain x-2y+6=0;

1,0) to x-2y+6=0 is d=5 5=5

Using the Pythagorean theorem, the chord length is 2* (13-5)=4 2

x2+y2-4=0 center of the circle(0,0)r=2

x2＋y2－4x＋4y－12＝0

Obtain: (x-2) 2+(y+2) 2=4 The center of the circle (2,-2)r=2 Let the common chord length be 2l, and the distance between the center of the circle is 2d, then there is:

d=√(2^2+2^2)/2=√2

l 2=r 2-d 2 Substituting the data yields:

l= 2 so the common chord length is 2 2

The two intersections are a(x1,y1),b(x2,y2)x2+y2-4=0 (1).

x2 y2 4x 4y 12 0 (2) do the difference to get y=x+2 (3) substitute (3) into (1) to get 2x 2+4x=0, solve x1=0, x2=-2 to get y1=2, y2=0

ab|= 2 root number 2

So the common chord length 2 root number 2

Lianli x2+y2-4=0 and x2+y2-4x+4y-12 0, obtain:

x1=0,y1=2

x2=-2,y2=0

Common chord length of two circles = [(x1-x2) 2+(y1-y2) 2] = =2 2

Subtract the two formulas to obtain: 4x+12y=40 =>x+3y-10=0 ......i.e. common chords.

The monolithic is: (x-5) 2+(y-5) 2=50 circle center o(5,5) r=5 2

o Distance to common chord d=(5+3*5-10) 1 2+3 2 = 10Common chord length = (r 2-d 2) *2=4 10Check it yourself, I don't know if there is a miscalculation.

Two equations are concatenated, find the two solutions of the system of equations, which are the coordinates of the two intersection points, and then find the distance between the two points.

Because the circle x 2+y 2=4 has a radius of 2 and the common chord length is 2 times the root number 3, the circle x 2+y 2=4 has a distance of 1 from the center of the circle to the common chord

Because the common chord of the two circles is x 2+y 2-4-(x 2+y 2+2ay-6) i.e. y=1 a

The center of the circle x 2 + y 2 = 4 is the origin.

So 1 a=1

Because x 2 + y 2 = 2 2 is a circle with the center of the circle at the origin and a radius of 2.

x 2+(y-2) 2=4 2 is a circle with the center of (0,2) and a radius of 4.

The big circle wraps the small circle, and the two circles are tangent.

No intersecting, no common chord, or common chord is 0

Solution: Two-circle equation:

x²+y²=4

x²+y²+4x+4y=12

Subtract the above two equations to get the equation of the straight line where the common chord is located:

x+y-2=0

It is easy to see that the length of the chord truncated by the circle of the line: x +y = 4 is the length of the common chord.

The center of the circle o(0,0), radius r=2

The chord centroid distance d= 2, (i.e., the distance from the center of the circle o(0,0) to the straight line x+y-2=0) Let the chord length be x, which can be obtained from the perpendicular diameter theorem and the Pythagorean theorem

r²=d²+(x/2)²

x/2=√(r²-d²)=√2

x=2 2, i.e. common chord length = 2 2

Upstairs copied, so a little bit of a small error, is now corrected as follows:

Let the common chord be ab

x²+y²-4=0 ……

x²+y²+4x+4y-12=0 ……Gotta :

4x-4y+8=0 ……

That is: x-y+2=0, which is the straight line where the common chord is.

The center of the circle x + y -4 = 0 is o(0,0) and the radius r = 2 is the distance from the center of the circle o to the common chord .

d=2/√2=√2

ab|/2=√(r²-d²)=√2

ab|=2√2

Let the common chord be ab

x²+y²-4=0 ……

x²+y²-4x+4y-12=0 ……Gotta :

4x-4y+8=0 ……

That is: x-y+2=0, which is the straight line where the common chord is. (If you don't understand, see the notes below).

The center of the circle x + y -4 = 0 is o(0,0) and the radius r = 2 is the distance from the center of the circle o to the common chord .

d=2/√2=√2

ab|/2=√(r²-d²)=√2

ab|=2√2

Note: Let the root of the system of equations be a(x1,y1) and b(x2,y2), and a can be substituted into , .

Then use 4x1-4y1+8=0, that is, x1-y1+2=0, and x2-y2+2=0 can be obtained in the same way

a(x1,y1),b(x2,y2) satisfy the equation x-y+2=0ab on the straight line x-y+2=0.

Subtract the equation of the two circles to obtain the equation of the straight line where the common chord is located (the equation is true when the common chord exists, and I don't know what the equation is obtained when it does not exist) The equation of the straight line where the common chord is located is y x 2, and the length of the common chord is 2 times the number 2,