The minimum distance between the point P (1,2) and the point on the circle C: x ^ 2 + y ^ 2 + 2kx + 2kx + K ^ 2 = 0 is What is the minimum distance between the point P (1,2) and the point on the circle C: x ^ 2 + y ^ 2 + 2kx + 2Y + K ^ 2 = 0

The minimum distance between the point P (1,2) and the point on the circle C: x ^ 2 + y ^ 2 + 2kx + 2kx + K ^ 2 = 0 is What is the minimum distance between the point P (1,2) and the point on the circle C: x ^ 2 + y ^ 2 + 2kx + 2Y + K ^ 2 = 0

Circle C: (x ^ 2 + 2kx + K ^ 2) + (y ^ 2 + 2Y + 1) = 1
===> (x+k)^2+(y+1)^2=1
It represents a circle whose center is on (- K, - 1) and radius is 1
The center of the circle moves arbitrarily on the straight line y - = - 1, and the circle is tangent to the X axis
Obviously, when the center of the circle is located at the intersection of x = 1 and y = - 1, that is, (1, - 1), the distance is the smallest
Minimum = 2 - (- 1) = 3

When k is the value, the curve represented by the equation x ^ 2 + y ^ 2-2kx + 2y-2k-2 = 0 is a circle, and the center coordinates of the circle with the largest area are obtained

Formula: (x-k) ^ 2 + (y + 1) ^ 2 = k ^ 2 + 2K + 3
That is, (x-k) ^ 2 + (y + 1) ^ 2 = (K + 1) ^ 2 + 2
So whatever the value of K, the equation represents a circle
The center of the circle is (k, - 1), and the radius is √ [(K + 1) ^ 2 + 2]
The radius has no maximum value, but only the minimum value √ 2, where k = - 1 and the center of the circle is (- 1, - 1)

It is known that circle C1: x ^ 2 + y ^ 2-2kx + K ^ 2-1 = 0 and circle C2: x ^ 2 + y ^ 2-2 (K + 1) y + K ^ 2 + 2K = 0
When the distance between the centers of their circles is the shortest, what is the position relationship___
I asked this question a few days ago. You said that we should use the distance formula
C1：（x-k）^2+y^2=1
C2：x^2+[y-(k+1)]^2=1
What is the distance? Please be more specific

Formula:
x²+y²-2kx+k²-1=0 (x-k)²+y²=1
Center coordinates: x = k, y = 0
x²+y²-2(k+1)y+k²+2k=0 x²+y²-2(k+1)y+(k+1)²=1 x²+[y-(k+1)]²=1
Center coordinates: x = 0, y = K + 1
Center distance = √ [(k-0) & # 178; + (0-k-1) & # 178;] = √ (2k & # 178; + 2K + 1) = √ [2 (K + 1 / 2) & # 178; + 1 / 2]
When k = - 1 / 2, the minimum distance between the two centers is √ 2 / 2
√2/2

Given that circle C is circumscribed with two circles x ^ 2 + (y + 6) ^ 2 = 1, x ^ 2 + (Y-2) ^ 2 = 1, the trajectory equation of the center of circle C is l, let l be the shortest distance between the point and m (x, y)
The small value is m, the distance between the point F (0,1) and the point m is n. find the trajectory equation L: find the equation of the trajectory Q of the point m satisfying M = n

The centers of two circles x ^ 2 + (y + 6) ^ 2 = 1 and x ^ 2 + (Y-2) ^ 2 = 1 are as follows:
C1 (0, - 6); C2 (0,2), radius 1;
So there is: | CC1 | - 1 = | CC2 | - 1; that is, CC1 = CC2
So: the trajectory l of point C is the vertical bisector of C1C2: y = - 2;
The trajectory equation L: y = - 2;
Then: M = | y + 2 |; n = | MF|
M = n, so: (y + 2) ^ 2 = x ^ 2 + (Y-1) ^ 2; simplify the equation of the trajectory Q of point m: x ^ 2 = 6y + 3