(1)k=2,則y=kx+1=2x+1,直線l：y=2x+1與雙曲線C：3x^2-y^2=1相交于不同的兩點(diǎn)A、B,故設(shè)A(x1,2x1 +1)、B(x2,2x2 +1).
將y=2x+1代入3x^2-y^2=1得：x^2+4x+2=0,由此得x1+x2=-4,(x1)(x2)=2,于是
|AB|=根號(hào){(x2-x1)^2+[(2x2 +1)-(2x1 +1)]^2}=根號(hào)[5*(x2-x1)^2]
=根號(hào){5*[(x2+x1)^2-4(x1)(x2)]}=2倍根號(hào)10；
(2)由（1）得A(x1,kx1 +1)、B(x2,kx2 +1)
將y=kx+1代入3x^2-y^2=1得：(3-k^2)x^2-2kx-2=0,由此得
x1+x2=(2k)/(3-k^2),(x1)(x2)=2/(k^2-3),判別式=(-2k)^2+8(3-k^2)=24-k^2>0,于是
|AB|=根號(hào){(x2-x1)^2+[(kx2 +1)-(kx1 +1)]^2}=根號(hào)[(1+k^2)*(x2-x1)^2]
=根號(hào){(1+k^2)*[(x2+x1)^2-4(x1)(x2)]}=2倍根號(hào){[(6-k^2)(1+k^2)]/[(3-k^2)^2]}；
因此以AB為直徑的圓的半徑為根號(hào){[(6-k^2)(1+k^2)]/[(3-k^2)^2]},圓心為(k/(3-k^2),3/(3-k^2)),故該圓的方程為：[x-k/(3-k^2)]^2+[y-3/(3-k^2)]^2=[(6-k^2)(1+k^2)]/[(3-k^2)^2]
如果坐標(biāo)原點(diǎn)在此圓上,則[k/(3-k^2)]^2+[3/(3-k^2)]^2=[(6-k^2)(1+k^2)]/[(3-k^2)^2]
即k^2=3或k^2=1
綜上,存在存在實(shí)數(shù)k,使得以線段AB為直徑的圓經(jīng)過(guò)坐標(biāo)原點(diǎn).請(qǐng)問(wèn)3樓主,要除去K^2不等于3嗎?是的，要舍去。