based on the

relationship between central and inscribed angles, the figure below show the relationship between angle

*ADB* (

θ) and angle

*AOB* (2

θ) and the obtuse angle

*BCD* (

α) and reflex angle

*BOD* (2

α).

From the figure:

$2\alpha = 2\theta + 180^\circ$

$\alpha = \theta + 90^\circ$

From triangle *BCD* (Using Cosine Law):

$x^2 = 10^2 + 12^2 - 2(10)(12)\cos \alpha$

$x^2 = 244 - 240\cos \alpha$

$x^2 = 244 - 240\cos (\theta + 90^\circ)$

$x^2 = 244 - 240(\cos \theta \cos 90^\circ - \sin \theta \sin 90^\circ)$

$x^2 = 244 + 240\sin \theta$ → Equation (1)

From right triangle *ABD*:

$x^2 + 8^2 = (2r)^2$

$x^2 = 4r^2 - 64$

$\sin \theta = \dfrac{8}{2r}$

$\sin \theta = \dfrac{4}{r}$

From Equation (1)

$4r^2 - 64 = 244 + 240\left( \dfrac{4}{r} \right)$

$4r^2 - 308 - \dfrac{960}{r} = 0$

$r^3 - 77r - 240 = 0$ → Equation (2)

$r = 10.0446, \,\, -3.8691, \,\, -6.1755$

Use *r* = 10.0446 cm *answer*