( 1)

y=kx+b

3x2-y2= 1

Eliminate x and y respectively.

(3-k^2)x^2-2kbx+(b^2- 1)=0

(3-k^2)y^2-6by+(3b^2-k^2)=0

The solution of the equation is the coordinates of two points AB, which is obtained by Vieta theorem.

x 1+x2=2kb/(3-k^2)

x 1x2=(b^2- 1)/(3-k^2)

y 1+y2=6b/(3+k^2)

y 1y2=(3b^2-k^2)/(3-k^2)

ab^2=(y2-y 1)^2+(x2-x 1)^2

=(y2+y 1)^2-4y 1y2+(x2+x 1)^2-4x 1x2

Substitute the above into the prescription to get AB.

(2)

The midpoint of AB is the center of the circle. If the circle passes through the origin, the distance from the origin to the midpoint is equal to half of AB.

According to the midpoint formula

x0=(x 1+x2)/2，y0=(y 1+y2)/2

x0^2+y0^2=[( 1/2)AB]^2

(x 1+x2)^2+(y 1+y2)^2=ab^2

Substituting numerical values, if K has a solution, there is a real number K. Let the circle of the diameter of line AB pass through the coordinate origin, and K can be obtained.

The train of thought of this problem is as follows

( 1)

y=kx+b

3x2-y2= 1

Eliminate x and y respectively.

(3-k^2)x^2-2kbx+(b^2- 1)=0

(3-k^2)y^2-6by+(3b^2-k^2)=0

The solution of the equation is the coordinates of two points AB, which is obtained by Vieta theorem.

x 1+x2=2kb/(3-k^2)

x 1x2=(b^2- 1)/(3-k^2)

y 1+y2=6b/(3+k^2)

y 1y2=(3b^2-k^2)/(3-k^2)

ab^2=(y2-y 1)^2+(x2-x 1)^2

=(y2+y 1)^2-4y 1y2+(x2+x 1)^2-4x 1x2

Substitute the above into the prescription to get AB.

(2)

The midpoint of AB is the center of the circle. If the circle passes through the origin, the distance from the origin to the midpoint is equal to half of AB.

According to the midpoint formula

x0=(x 1+x2)/2，y0=(y 1+y2)/2

x0^2+y0^2=[( 1/2)AB]^2

(x 1+x2)^2+(y 1+y2)^2=ab^2

Substituting numerical values, if K has a solution, there is a real number K. Let the circle of the diameter of line AB pass through the coordinate origin, and K can be obtained.

I hope you are satisfied ~