If the lengths of tangents drawn from a point $P$ to the circles $x^2+y^2-8x+40=0$,$5x^2+5y^2-25x+80=0$,and $x^2+y^2-8x+16y+160=0$ are equal,then the point $P$ is:

  • A
    $\left(8, \frac{15}{2}\right)$
  • B
    $\left(-8, \frac{15}{2}\right)$
  • C
    $\left(8, -\frac{15}{2}\right)$
  • D
    $\left(-8, -\frac{15}{2}\right)$

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