Find the product of the segments $BM.BN$ of the circles below

Question

Consider two circles with centers $A$ and $B$ and radii a and b, respectively, with $B$ lying on the circle with center $A$. If $MN$ is a chord of the circle with center $A$, tangent to the circle with center $B$, the product $BM.BN$ is:(Answer:$2ab$)

I solved it for a specific case. I needed the resolution for a general case where the point of tangency is not on the extension line of the two centers

$MO=NO=x\\ BO.CO = MO.NO \implies b.(2a-b)=x^2(I)$

$\triangle BOM:BM^2 = OM^2+OB^2=x^2+b^2(II)$

$(I)in(II): BM^2 = 2ab-b^2+b^2=2ab$

$BM=BN \therefore \boxed{BM.BN = 2ab}$

?????

Answer 1

Angles $\angle OMB$ and $\angle NCB$ are subtended by the same chord $BN$. Therefore, $$\triangle MOB \sim \triangle CNB$$ (AAA criterion) and the thesis follows immediately.

Answer 2

By Sine Rule,

$$\frac{BM}{\sin \angle BNO}=2a $$

$$BM=2a \times \sin \angle BNO \tag{1} $$

In $\Delta BON,$ $$\sin \angle BNO=\frac{BO}{BN}=\frac{b}{BN} \tag{2}$$

$(1), (2) \implies$

$$BM=2a \times \frac{b}{BN}$$ $$BM \times BN=2ab$$