When two circles intersect, there will be two common points, which exist in two original equations, and the coordinates of these two points are the solution sets of the two circle equations, so the coordinates of the two intersection points satisfy the equation obtained by subtracting the two circles.

Two points can determine a straight line and are unique, so if two circles are subtracted, the common chord of the two circles will be obtained.

Extended data:

Linear equation of the common chord of two intersecting circles;

Ruoyuan c1:(x-a1) 2+(y-b1) 2 = r12 or x2+y2+d1x+e1y+f.

Circle C2: (x-a2) 2+(y-B2) 2 = R2 2 or x2+y2+D2x+E2y+F2=0.

Then the linear equation passing through the intersection of two circles is: (x-a1) 2+(y-b1) 2-(x-a2) 2-(y-B2) 2 = r12-R22 or (D65438.

This is a variant of "the common chord equation is obtained by subtracting two intersecting circular equations"

Let two circles be

x^2+y^2+c 1x+d 1y+e 1=0①

x^2+y^2+c2x+d2y+e2=0 ②

Subtract these two expressions.

(x^2+y^2+c 1x+d 1y+e 1)-(x^2+y^2+c2x+d2y+e2)=0③

This is a linear equation.

1, first prove that this straight line passes through the intersection of two circles.

Let the intersection point be (x0, y0) and satisfy ① ②.

So satisfy 3.

So the intersection point is on the straight line ③.

2. Because there is only one straight line at two intersections.

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