Problem description:

1. Given that the circle O: x 2+y 2 = 25 and the circle O1:x 2+y 2-4x-2y-20 = 0 intersect at A and B, find the chord length AB.

2. It is known that far C is circumscribed by the circle X 2+Y 2-2x = 0 and tangent to the straight line X+sqrt(3)Y=0 at point Q(3, -sqrt(3)), and the equation of circle C is found.

3.a is a circle x 2+y 2-6mx-2 (m-1) y+10m 2-2m-24 = 0 (m belongs to r).

(1) proves that no matter what the value is, the center of the circle is on the same straight line L.

(2) Among the lines parallel to L, the lines that intersect, tangent and separate from the circle.

Do as much as you can, and then give points when you have done it all! !

Analysis:

1 Subtract the equations of two circles to get the equation of chord: 4X+2Y-5=0.

Distance from center (0,0) x 2+y 2 = 25 to chord 4X+2Y=5:

D=|-5|/SQRT(20)=SQR(5)/2

Half of the chord = sqr (5 2-5/4) = sqr (95)/2

So * * * Chord =SQR(95)

2 the center c is on the straight line passing through point q and on the straight line x+sqr (3) y = 0: y+sqr (3) = sqr (3) * (x-3).

At the same time, the distance from the center C to the straight line is equal to the distance from the circumscribed circle O( 1, 0) to the center minus 1, that is, sqr [(x-1) 2+y 2]-1= sqr [(x-3) 2+(.

Get C(0, -4*SQR(3)) or c (4,0).

So the equation: (x-4) 2+y 2 = 9.

X^2+(Y+4*SQR(3))^2=36

3 into the standard equation: (x-3m) 2+(y-m+ 1) 2 = 25.

Center of the circle (3M, M- 1)

X=3M，Y=M- 1。 If m is removed, X-3Y-3=0.

That is, the center of the circle is always on a straight line: X-3Y-3=0.

The linear equation parallel to the straight line X-3Y-3=0 is X-3Y+D=0.

Tangent when D=-3+5*SQR( 10) or -3-5*SQR( 10).

When it is between two numbers, it intersects, and when it is outside two numbers, it separates.

In doubt: Tel: 0371-* * * * * * * QQ: * * * * * * * *