If $$y = m_1 x + c_1$$ and $$y = m_2 x + c_2$$, $$m_1 \neq m_2$$ are two common tangents of circle $$x^2 + y^2 = 2$$ and parabola $$y^2 = x$$, then the value of $$8|m_1 m_2|$$ is equal to

We need to find $$8|m_1 m_2|$$ where $$y = m_1x + c_1$$ and $$y = m_2x + c_2$$ are the common tangents to the circle $$x^2 + y^2 = 2$$ and the parabola $$y^2 = x$$.

The parabola can be expressed as $$y^2 = 4 \cdot \tfrac{1}{4} \cdot x$$ with $$a = \tfrac{1}{4}$$, so its tangent in slope form is $$y = mx + \frac{1}{4m}$$.

For this line to be tangent to the circle $$x^2 + y^2 = 2$$, its perpendicular distance from the origin must equal $$\sqrt{2}$$. Hence,
$$\frac{\bigl|1/(4m)\bigr|}{\sqrt{1 + m^2}} = \sqrt{2}$$
and squaring both sides gives
$$\frac{1}{16m^2} = 2(1 + m^2)$$
so
$$1 = 32m^2 + 32m^4$$.

Solving the equation $$32m^4 + 32m^2 - 1 = 0$$ by setting $$u = m^2$$ leads to
$$32u^2 + 32u - 1 = 0$$
which yields
$$u = \frac{-32 \pm \sqrt{1024 + 128}}{64} = \frac{-32 \pm \sqrt{1152}}{64} = \frac{-32 \pm 24\sqrt{2}}{64}$$. Since $$u = m^2 > 0$$, we select the positive root to obtain
$$u = \frac{-32 + 24\sqrt{2}}{64} = \frac{-4 + 3\sqrt{2}}{8}$$.

Because both tangent lines correspond to $$m^2 = \frac{3\sqrt{2} - 4}{8}$$, it follows that $$m_1 = -m_2$$ and thus
$$|m_1 m_2| = m^2 = \frac{3\sqrt{2} - 4}{8}$$.

Finally, multiplying by 8 gives
$$8|m_1 m_2| = 8 \times \frac{3\sqrt{2} - 4}{8} = 3\sqrt{2} - 4$$.

Option A: $$3\sqrt{2} - 4$$