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

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.

• What do you mean by 'Find the distance from the flat surface to the point of tangency between the circles'? – Toby Mak Feb 26 at 11:20
• I would suggest you draw the situation, and draw it well! Then start drawing auxiliary lines, triangles that show up ... – Matti P. Feb 26 at 11:56

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 $$\frac{10}{3}.$$

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 ,$$HI = (HP+PI) = (\frac{4}{3}+2) = \frac{10}{3} \approx 3.33$$

• Minor issue: $AC=12$, so, in fact, $HP=\frac{4}{3}$ and $HI=\frac{10}{3}.$ – Patricio Feb 26 at 15:05
• @Patricio You are right. I corrected my mistake. Thank you for showing that. – Anirban Niloy Feb 26 at 16:04