已知\(F_1,F_2\)分别是双曲线\(C:\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\) \((a>0,b>0)\)的左,右焦点,过\(F_1(-2,0)\)的直线与双曲线\(C\)的左支交于\(A,B\)两点,若\(D\left(-\dfrac{1}{2},\dfrac{5}{8}\right)\)为\(\triangle ABF_2\)的内切圆的圆心,则该内切圆的半径为\((\qquad)\)
\(\mathrm{A}.\dfrac{\sqrt{41}}{8}\) \(\mathrm{B}.\qquad\dfrac{3}{2}\) \(\mathrm{C}.\qquad\dfrac{13}{8}\) \(\mathrm{D}.\qquad\dfrac{5\sqrt{17}}{8}\)
解析: 过\(D\)作\(DH\perp AB\),垂足为\(H\),则$$
|AF_2|-|BF_2|=|AH|-|BH|.$$又$$
\begin{cases}
& |AF_2|=|AF_1|+2a,\
& |BF_2|=|BF_1|+2a,
\end{cases}
\[所以
\]
|AF_2|-|BF_2|=|AF_1|-|BF_1|,$$于是$$|AH|-|BH|=|AF_1|-|BF_1|.$$因此\(H\)与\(F_1\)重合,因此所求内切圆半径即$$|DF_1|=\sqrt{
\left(-\dfrac{1}{2}+2\right)2+\left(\dfrac{5}{8}-0\right)2}=\dfrac{13}{8}.$$
