The moving circle passes through point a (2,0) and is tangent to the square of circle (x + 2) + the square of y = 4. The orbit equation of the center of the moving circle is obtained

The moving circle passes through point a (2,0) and is tangent to the square of circle (x + 2) + the square of y = 4. The orbit equation of the center of the moving circle is obtained

The distance difference between the center O and a (2,0), O '(- 2,0) = 2
The hyperbola (x ^ 2) / 1 - (y ^ 2) / 3 = 1 is obtained

Given the circle C: x ^ 2 + y ^ 2-2ax-2 (2a-1) y + 4 (A-1) = 0, when a changes, find the trajectory equation of the center of the circle

Let the coordinate of point m be (x, y) ∵ the circle m is tangent to the y-axis, the radius of circle m be | x | and the circle m and C: (x-a) & amp; sup8; + Y & amp; sup8; = A & amp; sup8; in conclusion, the trajectory equation of point m is Y & amp; sup8; = 8ax (X & gt; 1), y = 1 (X & lt; 1)

If a ∈ R, then the trajectory equation of the center of the moving circle x ^ 2 + y ^ 2-2ax-4ay + 5A ^ 2-1 = 0 is

The standard formula is: (x-a) ^ 2 + (y-2a) ^ 2 = 1
So the center of the circle is (a, 2a)
So the trajectory equation of the center of a circle is a straight line: y = 2x