For reference: If, $T$ and $K$ are points of tangency, measure of the $\overset{\LARGE{\frown}}{AB}$ is twice the measure of the $\angle ASL$, $BS=3$, $KS=1$ and $\frac{TB}{4} =\frac{AT}{3}$ .Calculate the area ratio of the triangular regions $ATK$ and $LKS$ . (Answer: $\frac{9}{4}$)
$ \frac{S_{\triangle KLS}}{S_{\triangle AKT}} = \frac{KL.KS}{AK.KT}\\ \therefore \frac{S_{\triangle KLS}}{ S_{\triangle AKT}}=\frac{KL}{AK.KT}\\ BK.AK = LK.KT\implies 4AK = KL.KT\\ \therefore \frac{S_{\triangle KLS}}{ S_{\triangle AKT}}=\frac{4KL}{KL.KT.KT}=\frac{4}{KT^2}$
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Claim 1 : $KT$ is the angle bisector of $\angle ATB$
Construction:
Draw $DT$ tangent to the circles at point $T$
Let $AT$ intersect smaller circle at $C$. Draw $CK$.
Proof: Using Alternate-segment theorem we get :
$$\angle CKA = \angle CTK..........(1)$$ $$\angle DTC = \angle CKT = \angle ABT..........(2)$$
Now, In $\triangle ATB$
$$\angle AKT = \angle KTB + \angle KBT ......(3)$$
From equations $(2)$ and $(3)$
We get, $$\angle CKA = \angle KTB....(4)$$
From equations $(1)$ and $(4)$, we get our result.
In $\triangle ATB$ , if we apply Angle-bisector theorem
we get,
$$\frac{TB}{AT} = \frac{BK}{AK}$$
Therefore, AK = 3$
Now, By the secant property of circles
We get, $LK.KT = 12$
Now, its trivial to prove that
$BS = LS = LK = 3$ and $KT$ = $4$ {$\angle AKS = \angle ATB = \frac{1}{2} \angle AOB$}
Therefore, $$\frac{[LKS]}{[ATK]} = \frac{1}{4}$$