If the equation x square + y square + 2mx-2y + m square + 5m = 0 represents a circle, find the value range of real number m? Who knows?

If the equation x square + y square + 2mx-2y + m square + 5m = 0 represents a circle, find the value range of real number m? Who knows?

The formula is (x + m) ^ 2 + (Y-1) ^ 2 = 1-5m,
Since the equation represents a circle, 1-5m > 0,
The solution is m

If circle x2 + y2-2mx + M2-4 = 0 is tangent to circle x2 + Y2 + 2x-4my + 4m2-8 = 0, find the set of all values of real number M

The two circles are transformed into the standard equation: (x-m) 2 + y2 = 4, (x + 1) 2 + (y-2m) 2 = 9, the coordinates of the center of the circle are a (m, 0) and B (- 1, 2m), and the radius is 2 and 3 respectively. From the tangency of the two circles, we get | ab | = 3 + 2 or | ab | = 3-2, that is, (M + 1) 2 + (0 − 2m) 2 = 5 or (M + 1) 2 + (0 − 2m) 2 = 1

If circle x2 + y2-2mx + M2-4 = 0 is tangent to circle x2 + Y2 + 2x-4my + 4m2-8 = 0, find the set of all values of real number M

The two circles are transformed into the standard equation: (x-m) 2 + y2 = 4, (x + 1) 2 + (y-2m) 2 = 9, the coordinates of the center of the circle are a (m, 0) and B (- 1, 2m), and the radius is 2 and 3 respectively. From the tangency of the two circles, we can get | ab | = 3 + 2 or | ab | = 3-2, that is, (M + 1) 2 + (0 − 2m) 2 = 5 or (M + 1) 2 + (0 − 2m) 2 = 1, we can get: (5m + 12) (m-2) = 0 or m (5m + 2) = 0, and the solution is m = - 125 or 2 or 0 or - 25 The set of all values is {- 125, - 25, 0, 2}