The trajectory equation of the center of the moving circle x + Y - (4m + 2) x-2my + 4m + 4m + 1 = 0 is? Moving circle x square + y square - (4m + 2) x-2my + 4m + 1 = 0 The formula is [x - (2m + 1)] and#178; + (y-m) ² = (2m + 1) ² + m and#178; - (4m + 1) ∴ [x-(2m+1)]²+(y-m)²=5m²,∴ m≠0 Let the center of the circle be (x, y) Then x = 2m + 1, y = M Eliminate M, We get x = 2Y + 1 and Y ≠ 0 The trajectory equation of the center of a circle is x-2y-1 = 0 (Y ≠ 0). Is there any other method besides this method?

The trajectory equation of the center of the moving circle x + Y - (4m + 2) x-2my + 4m + 4m + 1 = 0 is? Moving circle x square + y square - (4m + 2) x-2my + 4m + 1 = 0 The formula is [x - (2m + 1)] and#178; + (y-m) ² = (2m + 1) ² + m and#178; - (4m + 1) ∴ [x-(2m+1)]²+(y-m)²=5m²,∴ m≠0 Let the center of the circle be (x, y) Then x = 2m + 1, y = M Eliminate M, We get x = 2Y + 1 and Y ≠ 0 The trajectory equation of the center of a circle is x-2y-1 = 0 (Y ≠ 0). Is there any other method besides this method?

This method is a method to solve the trajectory equation: parameter method,
This is the way,
The abscissa and ordinate of the center of the circle are expressed by m, and the parametric equation is obtained
There is no other way

Let the equation x2 + y2-2 (M + 3) X-2 (1-4m2) y + 16m4 + 9 = 0. If the equation represents a circle, find the value range of M

If the equation of circle is [x - (M + 3)] 2 + [y - (1-4m2)] 2 = 1 + 6m-7m2, then 1 + 6m-7m2 ＞ 0, and the solution is - 17 ＜ m ＜ 1, so the value range of M is (- 17, 1)

X2 + y2-2 (M + 3) + 2 (1-4m2) y + 16m4 + 9 = 0 is a circle
Ask for detailed explanation
The answer is 0

[x - (M + 3)] ^ 2 + [y + (1-4m ^ 2)] ^ 2 = - 16m ^ 4-9 + (M + 3) ^ 2 + (1-4m ^ 2) ^ 2, so R ^ 2 = - 16m ^ 4-9 + (x + 3) ^ 2 + (1-4m ^ 2) ^ 2 > 0-16m ^ 4-9 + (M + 3) ^ 2 + (1-4m ^ 2) ^ 2 = - 16m ^ 4-9 + m ^ 2 + 6m + 9 + 16m ^ 4-8M ^ 2 + 1 = - 7m ^ 2 + 6m + 1 > 07m ^ 2-6m-1

Given the circle equation x2 + y2-2 √ m + 1x - √ my + m + 1 = 0, find the trajectory equation of the center of the circle

First, we can get [x - √ (M + 1)] ^ 2 + [y - (√ m) / 2] ^ 2 = m / 4
Radius = (√ m) / 2. So m > 0
Let the center of the circle be (x, y)
X = √ (M + 1) so x ^ 2 = m + 1
Y = (√ m) / 2, so y ^ 2 = m / 4
So x ^ 2 = 4Y ^ 2 + 1
So the trajectory is x ^ 2-4y ^ 2 = 1
Because m > 0. So x > 1. Y > 0