In the adjoining figure $TP$ and $T'Q$ are parallel tangents to a circle of radius $r$, with $T$ and $T'$ the points of tangency. $PT''Q$ is a third tangent with $T'''$ as a point of tangency. If $TP=4$ and $T'Q=9$ then $r$ is
[asy] unitsize(45); pair O = (0,0); pair T = dir(90); pair T1 = dir(270); pair T2 = dir(25); pair P = (.61,1); pair Q = (1.61, -1); draw(unitcircle); dot(O); label("O",O,W); label("T",T,N); label("T'",T1,S); label("T''",T2,NE); label("P",P,NE); label("Q",Q,S); draw(O--T2); label("$r$",midpoint(O--T2),NW); draw(T--P); label("4",midpoint(T--P),N); draw(T1--Q); label("9",midpoint(T1--Q),S); draw(P--Q);[/asy]
$\mathrm{(A)\ } 25/6 \qquad \mathrm{(B) \ } 6 \qquad \mathrm{(C) \  } 25/4 \qquad$
$\mathrm{(D) \  } \text{a number other than }25/6, 6, 25/4 \qquad$
$\mathrm{(E) \  }\text{not determinable from the given information}$
[asy] unitsize(45); pair O = (0,0); pair T = dir(90); pair T1 = dir(270); pair T2 = dir(25); pair P = (.61,1); pair Q = (1.61, -1); draw(unitcircle); dot(O); label("O",O,W); label("T",T,N); label("T'",T1,S); label("T''",T2,NE); label("P",P,NE); label("Q",Q,S); draw(T--P); label("4",midpoint(T--P),N); draw(T1--Q); draw(P--Q); draw(T--T1); pair R = (.61,-1); draw(P--R); label("R",R,S); label("4",midpoint(T1--R),S); label("5",midpoint(R--Q),S); label("r",midpoint(O--T),W); label("r",midpoint(O--T1),W);[/asy]
Drop the perpendicular from $P$ to $T'Q$ and let the foot be $R$. Note that $PTT'R$ is a rectangle. Also, from Two Tangents, $PT''=4$ and $QT''=9$, so $PQ=13$. Therefore, from the Pythagorean Theorem on $\triangle PRQ$, $PR=\sqrt{13^2-5^2}=12$. We now see that $r=\frac{PR}{2}=6, \boxed{\text{B}}$.