$y^{2}=4 x$
$2 y \frac{d y}{d x}=4$
$m_{\mathrm{T}}=\frac{2}{y}=\frac{2}{2}=1$
Circle $\rightarrow \mathrm{S}+\lambda \mathrm{L}=0$
$(\mathrm{x}+1)^{2}+(\mathrm{y}-\alpha)^{2}+\lambda(\mathrm{x}+1)=0$...(1)
differentiate
$$
\begin{array}{l}
2(x+1)+2(y-\alpha) \frac{d y}{d x}+\lambda=0 \\
x=1, y=2 \\
4+2(2-\alpha) m_{T}+\lambda=0 \\
\lambda=2 \alpha-8...(2)
\end{array}
$$
$(1,2)$ satisfies cq. $(1)$
$$
\begin{array}{l}
2^{2}+(2-\alpha)^{2}+2 \lambda=0 \\
\alpha^{2}-4 \alpha+8+2(2 \alpha-8)=0 \\
\alpha^{2}=8 \\
\alpha=2 \sqrt{2}
\end{array}
$$