What is the length of the radius of the circle

Consider BD. Using the cosine rule for $\triangle ABD$ and $\triangle BCD$ , we obtain:

${BD}^2 = {AB}^2 + {AD}^ 2 - 2.AB.AD\cos(\angle BAD)$

${BD}^2 = {BC}^2 + {CD}^ 2 - 2.BC.CD\cos(\angle BCD)$

Equating these, we obtain:

${AB}^2 + {AD}^ 2 - 2.AB.AD\cos(\angle BAD) = {BC}^2 + {CD}^ 2 - 2.BC.CD\cos(\angle BCD)$

However, ${AB}^2 + {AD}^ 2 = {BC}^2 + {CD}^ 2 = 85$ .

So:

$2.AB.AD\cos(\angle BAD) = 2.BC.CD\cos(\angle BCD)$

$AB.AD\cos(\angle BAD) = BC.CD\cos(\angle BCD)$

Since ABCD is a cyclic quadrilateral, $\angle BCD = 180^{\circ} - \angle BAD$ .

$18 \cos(\angle BAD) = 42 \cos(180^{\circ} - \angle BAD)$

In general, $\cos(\theta) = \cos(180^{\circ} - \theta)$ . So:

$18 \cos(\angle BAD) = 42 \cos(\angle BAD)$

$0 = (42 - 18) \cos(\angle BAD)$

$0 = \cos(\angle BAD)$

$\angle BAD = 90^{\circ}$ (as $\angle BAD$ lies between $0^{\circ}$ and $180^{\circ}$ )

Therefore, [BD] is a diameter of the circle.

$BD = \sqrt {85}$

This means that the radius, $r = \frac{\sqrt{85}}{2}$

Squaring the radius gives: $\frac{85}{4}$ = 21.25