$解：(2)分兩種情況$
$①當(dāng)AC經(jīng)過(guò)圓心，B_2C是⊙O的切線時(shí)$
$此時(shí)B_2C⊥B_2O$
$∵B_2(\frac {\sqrt{2}}{2}，－\frac {\sqrt{2}}{2})$
$∴y_{B_2O}=－x$
$∴直線B_2C的斜率為1$
$設(shè)直線B_2C的解析式為y=x+b$
$把(\frac {\sqrt{2}}{2}，－\frac {\sqrt{2}}{2})代入：$
$\frac {\sqrt{2}}{2}+b=－\frac {\sqrt{2}}{2}$
$b=－\sqrt{2}，∴y_{B_2C}=x－\sqrt{2}$
$當(dāng)y=0時(shí)，x－\sqrt{2}=0，x=\sqrt{2}$
$∴C(\sqrt{2}，0)$
$②當(dāng)B_2C經(jīng)過(guò)圓心，AC是⊙O的切線時(shí)$
$∵A(－1,0)∴C的橫坐標(biāo)為－1$
$由①知，直線B_2O的解析式為y=－x$
$當(dāng)x=－1時(shí)，y=1$
$∴C(－1，1)$
$綜上所述：C(\sqrt{2}，0)或(－1,1)$
