Let PA and PB be the tangent segments drawn from point P$(6, 8)$ to the circle with the centre at origin O. The radius of circle for which the area of quadrilateral PAOB is maximum, is...

Let the radius of the circle be $r$.

The centre of the circle is $O(0, 0)$ and the external point is $P(6, 8)$.

The distance between $O$ and $P$ is:

$OP = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

The length of the tangent segments $PA$ and $PB$ from $P$ to the circle is:

$PA = PB = \sqrt{OP^2 - r^2} = \sqrt{100 - r^2}$

The quadrilateral $PAOB$ consists of two congruent right-angled triangles, $\triangle OAP$ and $\triangle OBP$.

The area of the quadrilateral $PAOB$ is:

$A = 2 \times \text{Area of } \triangle OAP = 2 \times \left( \dfrac{1}{2} \times OA \times PA \right) = r \sqrt{100 - r^2}$

To maximize the area $A$, we can maximize its square, $A^2$:

$f(r) = A^2 = r^2 (100 - r^2) = 100r^2 - r^4$

Differentiating $f(r)$ with respect to $r$:

$f'(r) = 200r - 4r^3$

Setting the derivative to zero for maximum/minimum:

$200r - 4r^3 = 0$

$4r(50 - r^2) = 0$

Since $r > 0$, we have $50 - r^2 = 0 \Rightarrow r^2 = 50$

$r = \sqrt{50} = 5\sqrt{2}$

To verify it is a maximum, check the second derivative:

$f''(r) = 200 - 12r^2$

At $r = 5\sqrt{2}$, $f''(5\sqrt{2}) = 200 - 12(50) = 200 - 600 = -400 < 0$

Thus, the area is maximum when the radius is $5\sqrt{2}$.