Answer
$$y_1=\sqrt{3}x+1$$
$$y_2=- \sqrt{3}x+1$$
Work Step by Step
The general equation of a line is $y=mx+d$.
Since both tangent lines have the same $y$-intercept, $(0,1)$, we have $$y_1=m_1x+1, \\y_2=m_2x+1.$$To find the unknown parameters $m_1$ and $m_2$, we substitute $"y"$ in the equation of the circle with $"m_i x+1"$ and force the resultant quadratic equation to have only one root.
So we have $$x^2+(y+1)^2-1=0 \quad \Rightarrow \quad x^2+ (m_ix+2)^2-1 \\ =x^2+m_i^2x^2+4m_ix+4-1 \\=(1+m_i^2)x^2+4m_ix+3=0$$If we want that the above quadratic equation has only one root, its discriminant must be zero. So $$\Delta = b^2 - 4ac= 16m_i^2-12-12m_i^2=0 \\ \Rightarrow m_i= \pm \sqrt{3}$$
Hence, $$y_1=\sqrt{3}x+1,$$ $$y_2=- \sqrt{3}x+1.$$
