Yes but there are four possible constructions- two when externally tangent two when internally tangent.[Maybe!]SANZEED wrote:Can anyone confirm if this is the correct figure?

For each of which there are cases when $\omega$ touches $AB,AC$ on segments or on extensions.

So to avoid complications I would recommend using directed lenghs.

Since no one posted a solution, here's a little hint to shed some light:

Let $A$ be a point on $\odot \alpha$. A circle $\beta$ is tangent (internally or externally) to $\alpha$ with tangency point $B$. Let $\ell$ be the length of the tangent from $A$ to $\beta$. Let the radius of $\alpha,\beta$ be $R,r$ respectively. Prove that \[AB=\ell.\sqrt{\frac{R}{R \pm r}}\][The sign is negative when internally tangent and positive when externally tangent.]

Use the hint, the fact whether $K,\omega$ are externally or internally tangent or whether $\omega$ touches $AB,AC$ on segments or extension won't matter.