If a is a real number, then the linear equation of the center of the circle (x-a) 2 + (y + 2a) 2 = 1 is () A. 2x+y=0B. x+2y=0C. x-2y=0D. 2x-y=0

If a is a real number, then the linear equation of the center of the circle (x-a) 2 + (y + 2a) 2 = 1 is () A. 2x+y=0B. x+2y=0C. x-2y=0D. 2x-y=0

From the meaning of the title, the coordinate of the center of the circle is x = ay = - 2A, and the elimination of a leads to 2x + y = 0; so the linear equation of the center of the circle is 2x + y = 0, so a

It is known that the circle C: x ^ 2 + y ^ 2-2ax + 2 (2a-1) y + 4 (A-1) = 0, where a belongs to R
(1) Prove that circle C passes through fixed point
(2) When a changes, find the trajectory equation of the center of the circle
(3) The equation for finding the smallest circle C

(1) The equation can be simplified as: (x-a) ^ 2 + (y + A-2) ^ 2 = 2 (A-1) ^ 2 = (1-A) ^ 2 + (A-1) ^ 2. It is known from the above formula that when a ≠ 1, there is: when x = 1, y = 1, the above formula is constant. So the circle must pass the constant point (1,1). (2). From (1), the center O is: (- A, 2-A). And the circle passes the constant point (1,1). Then we can know the tangent of a (1,1)

Given the circle x ^ 2 + y ^ 2-2ax-4ay + 5A ^ 2-4 = 0. Find the radius of the circle, the center coordinates of the circle and the linear equation satisfied by the center coordinates of the circle

(x-a) ^ 2, y ^ 2 = 3A ^ 2 = (root 3 · a) ^ 2, so the radius of the center of the circle (a, 0) root 3 · a understand