It is known that the symmetric point of point a (2,3) on the circle about the straight line x + 2Y = 0 is still on the circle, and the chord length intersecting with the straight line X-Y + 1 = 0 is double root sign 2, so the equation of the circle is solved By the way, the distance between the lower chords

It is known that the symmetric point of point a (2,3) on the circle about the straight line x + 2Y = 0 is still on the circle, and the chord length intersecting with the straight line X-Y + 1 = 0 is double root sign 2, so the equation of the circle is solved By the way, the distance between the lower chords

(x-4) · (x-4) + (y + 1) · (y + 1) = 20 or (X-8) · (X-8) + (y + 1) · (y + 1) = 52, it is easy to find that the symmetry point of a is (2, - 5), then the center of the circle must be on y = (- 5 + 3) * 0.5 = - 1, let the center of the circle be (a, - 1), and the equation be (x-a) · (x-a) + (y + 1) · (y + 1) = R · R

It is known that the equation of circle C 'is x2 + (Y-1) 2 = 4, and the center coordinates of circle C are (2, - 1). If circle C and circle C' intersect at two points a and B, and ab = 2 √ 2, the equation of circle C is obtained

(x-2)2+(y+1)2=4

General equation of circle (15 17:55:25)
If the line 3x + 4y-12 = 0 intersects with X axis and Y axis with a and B respectively, then the equation of inscribed circle of △ AOB is (including process)

A（4,0）,B（0,3）
AO=4,BO=3,AB=5
（3+4+5）*R=3*4
Radius of inscribed circle r = 12 / 12 = 1
The center of the circle (1,1)
(x-1) square + (Y-1) square = 1

Let the minimum value of function f (x) = x2-2x + 2 & # 160; &# 160; &# 160; X ∈ [T, t + 1] x ∈ R (closed interval) be g (T), and find g (T)

It is reduced to: F (x) = (x-1) ^ 2 + 1
It can be seen that the opening of the function is upward, the stable axis is x = 1, and the vertex is (1,1)
When 1