(1) meter (5, t)

(2)

hand over

∴CB=CB"=OA=5,B"M=BM=4-t

From Pythagorean theorem,

OB"= the square of CB under the root sign-the square of Co =3.

∴AB"=5-3=2

From Pythagorean theorem,

The square of BM = the square of AB+the square of AM.

That is, the square of (4-t) = the square of 4+t.

∴t=3/2

∴M(5,3/2)

Let the analytical formula of straight line CM be: y = kx+b.

Introduce C (0 0,4) and M (5 5,3/2)

Then k =- 1/2 and b = 4.

∴y=- 1/2x+4

(3) existence. (fatal point)

①CO=OQ

∴OQ=4

QH⊥AO in h

Let Q(a,-1/2a+4)

Derived from Pythagorean theorem

The square of a +(- 1/2a+4) is 16.

A 1= 16/5, a2=0 (omitted)

∴q 1( 16/5, 12/5)

②CQ=OC

∴CQ=4

Make QI⊥OC in I through point Q.

Let Q(a,-1/2a+4)

Derived from Pythagorean theorem

The square of a +[4-(- 1/2a+4)] is the square of 16.

A 1 = 8 times the number of roots of 5, A2 =-8 times the number of roots of 5 (omitted)

∴ Q2 (8 root number 5, 4-5 root number 5)

③CQ=OQ

QJ⊥OC at point J.

Let Q(a,-1/2a+4)

∫QC = OQ

∴OJ=CJ= 1/2OC=2

∴- 1/2a+4=2

∴a=4

∴Q3(4,2)