If the radical axis of the circles $x^2+y^2+2 \alpha x+2 \beta y+c=0$ and $x^2+y^2+\frac{3}{2} x+4 y+c=0$ touches the circle $x^2+y^2+2 x+2 y+1=0$,then $4 \alpha \beta-8 \alpha-3 \beta+10=$

  • A
    $2$
  • B
    $-2$
  • C
    $4$
  • D
    $-4$

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