Finding the distance from the flat surface to the point of tangency between two circles with having radii $2$ and $10$ respectively

Question

I have had trouble with this problem lately.

Question:

Assume there are two circle with radii $2$ and $10$. The both circle are tangent to each other and are on a flat surface. Find the distance from the flat surface to the point of tangency between the circles.

Please consider the large radius and small radius with the mentioned information. Thanks, NJC

Answer 1

An alternative method: Let the flat surface be the $x$ axis and $(2,2)$ be the center of the small circunference. In order for the two circumferences to be tangent to each other, the big circumference has to have its center at $(2+\sqrt{80},10).$ As the point of tangency is on the line that connects the centers of the two circunferences, it needs to be a convex combination of them. In particular, as it is at a distance of $2$ from $(2,2)$ and a distance of $10$ from $(2+\sqrt{80},10)$ it should be

$$ \frac{10}{12}(2,2)+\frac{2}{12}(2+\sqrt{80},10)=(2+\frac{\sqrt{80}}{6},\frac{10}{3}),$$

so the distance to the $x$ axis is $\frac{10}{3}.$

Answer 2

Let $A$ be the radius of the small circle and $C$ be the radius of large circle. Connecting those necessary point, we get $AC = 12$, $GC = 8$ and $AB = PI = 2$. We need to find out the $|HI|$.

$\triangle ACG \sim \triangle AHP$ and so

$\frac{HP}{AH} = \frac{GC}{AC}$ $\implies HP = \frac{8×2}{12} \implies HP = \frac{4}{3}$

Hence, distance from the flat surface to the tangent point ,$HI = (HP+PI) = (\frac{4}{3}+2) = \frac{10}{3} \approx 3.33$