The trajectory equation of the center C of the moving circle is obtained when the moving circle passes through the fixed point a (- 2,0) and is tangent to the fixed circle (X-2) ^ 2 + y ^ 2 = 12 (1)

Let the center of the moving circle C (m, n) pass through the fixed point a (- 2,0), so that the equation of the moving circle (x-m) ^ 2 + (y-n) ^ 2 = (M + 2) ^ 2 + n ^ 2, the moving circle can only be circumscribed with the fixed circle, so the root ((m-2) ^ 2 + n ^ 2) = 2 root 3 + root ((M + 2) ^ 2 + n ^ 2) is a stupid method. There is a simple way to draw a picture and set the center of the circle BCB -

If the moving circle is tangent to the X axis and the chord length cut by the straight line y = x is 2, then the trajectory equation of the center of the moving circle is______ ．

Let (x, y) be the center coordinate of the circle, then the radius of the circle is | y | and the chord center distance is d = | X-Y | 2. Because the chord length is 2, there is y2 = 1 + (| X-Y | 2) 2, which is sorted out to be x2-y2-2xy + 2 = 0, so we should fill in x2-y2-2xy + 2 = 0

The trajectory equation of the center of a moving circle which is circumscribed with the circle C: x 2 + y 2-6x = O and tangent to the Y axis

Circle X & # 178; + Y & # 178; - 6x = 0, namely: (x-3) &# 178; + Y & # 178; = 9
The center of the circle (3,0), the radius is 3
The center of a circle tangent to both the circle and the y-axis may be on the x-axis or on a parabola
The trajectory equation is y = 0 or Y & # 178; = 12x

The trajectory equation of the center C of the moving circle is obtained when the moving circle passes through the fixed point a (- 2,0) and is tangent to the fixed circle (X-2) ^ 2 + y ^ 2 = 12 (1)