X 2+Y 2-2ax-2by+A 2- 1 = 0。 When A and B change, if circle B always bisects the circumference of circle A, find the trajectory equation of center B and the equation of the circle with the smallest radius of circle B..

Solution:

Circle a: x2+y2+2x+2y-2 = (x+1) 2+(y+1) 2 = 4.

Center (-1,-1), radius 2.

Circle b:

x^2+y^2-2ax-2by+a^2- 1=0

= = & gt(x-a)^2+(y-b)^2=b^2+ 1

Center (a, b), radius: radical sign (b 2+ 1)

x^2+y^2+2x+2y-2=0

x^2+y^2-2ax-2by+a^2- 1=0

Subtract both:

(2+2a)x+(2+2b)y= 1+a^2

Circle b always bisects the circumference of circle a. The center (-1,-1) of the circle A brings (-1,-1) into: (2+2a) x+(2+2b) y = 1+A 2 to get the circle B.

Then find the equation with the smallest radius.