Find the radius of the circle3

Drop a perpendicular from $O$ to $\overline {DB}$ to meet the later at $E$ .

Let $\overline {DB}$ intersects the circle at $C$ and $F$ respectively. Then

$8\times |\overline {FB}|=6^2\implies |\overline {FB}|=4.5,|\overline {CF}|=3.5,|\overline {CE}|=1.75$

$|\overline {OE}|^2=12^2-11.75^2\implies |\overline {OC}|^2=12^2-11.75^2+1.75^2=9$

Therefore the radius of the circle is

$|\overline {OC}|=\sqrt 9=\boxed 3$ .

Extend $OD$ to meet the circle at $G$ . $BC$ and $OD$ cut the circle at $E$ and $F$ respectively. Let the radius of the circle $FO=OG = r$ and $CE=d$ .

Using theorems, we can easily solve this problem.

• By tangent-secant theorem , we have $BC\cdot BE = AB^2 \implies 8(8-d) = 6^2 \implies d = 3.5$ .
• By two-secant theorem , $DF \cdot DG = CD \cdot DE \implies (12-r)(12+r) = 10(10+3.5) \implies 144-r^2 = 135 \implies r = \boxed 3$ .