(20 1 1? Shantou simulation) Let the circle Q cross point P (0 0,2), and the length of the truncated chord RG on the X axis is 4. (1) The equation for finding the locus e of the center q; (2)

Solution: (1) Let the coordinates of the center Q be (x, y), as shown in the figure, the center Q is Qh ⊥ and the x axis is h.

Then h is the midpoint of RG, and in Rt△RHQ, |QR|2=|QH|2+|RH|2.

∵|QR|=|QP|，|RH|=2，

∴x2+(y-2)2=y2+4

That is, x2=4y, so the equation of trajectory e is x2 = 4y. (5 points).

(2) let A(xA, yA), B(xB, yB), M(xm, ym), N(xN, yN).

The equation of straight line AB is y=kx+ 1(k≠0), and the simultaneous x2=4y is: x2-4kx-4=0.

∴xm=xa+xb2=2k,ym=kxm+ 1=2k2+ 1，

The coordinate of point m is (2k, 2k2+ 1). (7 points)

Similarly, the coordinate of point n is (? 2k，2k2+ 1)。

The slope of the straight line MN is kmn = ym? yNxM？ xN=k2？ 1k2k+ 1k=k2？ 1k，

Its equation is y? 2k2？ 1=k2？ 1k(x？ 2k), and k(y-3)=(k2- 1)x,

Regardless of the value of k, point (0,3) satisfies the equation.

∴ The straight line MN passes through the fixed point (0,3). (12)