|ab|=√((x2-x 1)？ +(y2-y 1)？ )

Y2 = kx2+ 1，y 1 = kx 1+ 1。

replace

|ab|=√((x2-x 1)？ +(kx2-kx 1)？ )

=√( 1+k？ )|x2-x 1|

Substitute the straight line l:y=kx+ 1 into the hyperbola C: 3x 2-Y 2 = 1.

get

3x？ -(kx+ 1)？ = 1

arrange

3x？ -k？ x？ -2kx-2=0

The absolute value of the difference between two roots is

|x2-x 1|=√((x 1+x2)？ -4x 1x2)=√(2k/(3-k？ ))? +8/(3-k？ ))

=√(2k+8(3-k？ )/|3-k？ |

=√(2k+24-8k？ )/|3-k？ |

|ab|=√( 1+k？ )*√(2k+24-8k？ )/|3-k？ |(2)a(x 1，kx 1

+ 1)、b(x2，kx2

+ 1)

Substituting y=kx+ 1 into 3x 2-y 2 = 1, we get: (3-k 2) x 2-2kx-2 = 0.

X 1+x2 = (2k)/(3-k 2), (x 1) (x2) = 2/(k 2-3), discriminant = (-2k) 2+8 (3-k 2) = 24-. 0, so

|ab|= radical sign {(x2-x 1) 2+[(kx2

+ 1)-(kx 1

+1)] 2} = radical sign [(1+k 2) * (x2-x1) 2]

= radical sign {(1+k2) * [(x2+x1) 2-4 (x1) (x2)]} = quadratic radical sign {[(6-k 2) (1+k)

So the radius of a circle with a diameter of ab is the radical sign {[(6-K2) (1+K2)]/[(3-K2) 2]}, and the center of the circle is (k/(3-k 2), 3/(3-k 2).

If the coordinate origin is on this circle, [k/(3-k2)] 2+[3/(3-k2)] 2 = [(6-k2) (1+k2)]/[(3-k2)]

That is, k 2 = 3 (disagree, give up) or k 2 = 1

To sum up, there is a real number k, which makes the circle of the diameter of the line segment ab pass through the coordinate origin.