Kick that altitude

First we need to observe that $\triangle BPQ$ and $triangle ABC$ are similar.

Now, using $\Delta = \frac{1}{2} ab \sin \angle C$ , we will have after some cancellation,

$\dfrac{[BPQ]}{[ABC]} = \dfrac{BQ \times BP}{ AB \times AC} = \dfrac{PQ^2}{b^2}$ [ as both the ratios are equal to $\frac{PQ}{ b}$ ]

or, $\dfrac{ 1}{9} = \dfrac{8}{b^2}$ and we will have $b = 6\sqrt{2}$ .

Now, $PQ$ is one of the sides of the orthic triangle and so $PQ = b \cos \angle B$

Therefore, $2\sqrt{2} = 6 \sqrt{2} \times \cos \angle B$

From where we wil get $\cos \angle B = \frac{1}{3}$ and so $\sin \angle B = \frac{2\sqrt{2}}{3}$

Lastly, from the laws of sines in triangle ABC, we have ,

$\dfrac{b}{\sin \angle B} = 2R$ . Putting the values and solving, we have $R = \dfrac{9}{2}$ and hence the answer.

The triangle PBQ is similar to the triangle ABC. Since the ratio of [ABC] to [PBQ] is 18/2=9, we have BC:BQ=AC:PQ=3. So $AC=6\sqrt{2}$ and $\cos{\angle{B}}=\frac{1}{3}$ . Hence $2R=\frac{AC}{\sin{\angle{B}}}=9$ .

Because $\angle AQC= \angle APC = 90^\circ,$ the points $A,$ $Q,$ $P,$ and $C$ lie on the circle with diameter $AC.$ By the inscribed angle theorem, $\angle CPQ = 180^\circ - \angle CAQ,$ so $\angle QPB = \angle BAC.$ Likewise, $\angle PQB = \angle ACB.$ Thus, the triangles $BAC$ and $BPQ$ are similar (with this order of vertices). The ratio of their areas is $18/2=9,$ so $|AC|=3\cdot |QP| = 6\sqrt{2}.$ Also, $\cos \angle ABC = \frac{|BP|}{|BA|}=\frac{1}{3}.$ Therefore, $\sin \angle ABC = \frac{2\sqrt{2}}{3}.$

By the Law of Sines, the radius of the circumscribed circle of $ABC$ equals $\frac{|AC|}{2\sin \angle ABC}= \frac{6\sqrt{2}}{2\cdot \frac{2\sqrt{2}}{3}}=\frac{9}{2}.$

$AQPC$ is a cyclic quadrilateral, angle $QPB$ equals angle $A$ . Therefore triangle $BQP$ is similar to triangle $BCA$ . Since the ratio of the area is 9, the ratio of the sides is 3. $PQ=2\sqrt{2}$ , so $AC=6\sqrt{2}, BQ=\frac{1}{3}BC, BP=\frac{1}{3}AB$ . Let $BC=a, AB=c$ , Using the Pythagorean theorem on $CAQ$ and $CBQ$ , we have the equation: $(6\sqrt{2})^2 - (c-1/3a)^2 = a^2 - (\frac{1}{3}a)^2$ , we simplifies to $a^2 + c^2 = 72 + \frac{2}{3}ac$ . Using Law of Cosine, we also have the equation: $(6\sqrt{2})^2 = a^2 + c^2 - 2ac Cos(B)$ . Combining the two equations give us: $CosB=1/3$ , so $SinB=\frac{2\sqrt{2}}{3}$ . Finally, using Law of Sine, $R = \frac{6\sqrt{2}}{2Sin(B)}=\frac{9}{2}$ .