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Imagine I have two identical circles of radius 1, placed side-by-side so they touch at one point, and a tangent line that touches both circles at one point each, like so:

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There's a vaguely triangular-shaped region defined by the lower-inward curves of the two circles and the line below them. It would be possible to place a circle inside that region that would touch the other two circles and the tangent line at one point each. What is the radius of this circle?

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  • $\begingroup$ This may look sort of like a homework question, but it's been years and years since I had math homework. I'm mostly just curious about how to work something like this out. $\endgroup$ – Mason Wheeler May 23 '16 at 12:22
  • $\begingroup$ In addition to direct approaches like those in the answers. this can be viewed as a special case of Descrates 4 circle theorem. If $r_1, r_2$ is the radius of the two circles on the sides and $r$ is the radius of the circle in between, one has $\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{2}{\sqrt{r_1 r_2}}$. In current case where $r_1 = r_2 = 1$, this leads to $r = \frac14$. $\endgroup$ – achille hui May 23 '16 at 12:58
  • $\begingroup$ A nice explanation is found in en.m.wikipedia.org/wiki/Ford_circle $\endgroup$ – Michael Hoppe May 23 '16 at 14:56
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Now you can use triangle area formula and heron's formula to find $x$.

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The figure defines a rectangle with the the given line as a side and centers of circles as two vertices. The center of your circle is on the vertical axis of symmetry, and it must be at some distance $r$ from the bottom line and $(R+r)$ from the big circle's center (where $R$ is the big circle radius and $r$ is your circle radius). That leads to an equation $$R^2 + (R-r)^2 = (R+r)^2$$

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enter image description here

Always draw a picture, and always add lines everywhere you can think of that might be relevant. If you think the drawing is enough, don't look below.

Assuming $ABCD$ is a rectangle, and setting the radius of the red circle to $r$, we get that $\triangle ADE$ is a right triangle with legs $1$ and $1-r$ and hypotenuse $1+r$. Now use the Pythagorean theorem.

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Just for convenience, place the left circle at $(-1,1)$ and the right circle at $(1,1)$. Then the center of the circle your describing is at $(0,r)$ where $r$ is the radius. Think about the right triangle formed from the center of one of the big circles with the the center of the small circle and a point directly below the big circle's center.
enter image description here

This forms a triangle where you can use the Pythagorean theorem to get $$ (1+r)^2=1^2+(1-r)^2 $$ which you can solve for $r$.

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Kiss Precise to Princess Elizabeth in Descartes letter...regarding Soddy Cirles:

$$ 2 \left(\frac{1}{r^2} +\frac{1}{r_1^2}+ \frac{1}{r_2^2} +\frac{1}{r_3^2}\right)= \left(\frac{1}{r} +\frac{1}{r_1}+ \frac{1}{r_2} +\frac{1}{r_3}\right)^2$$

$$ r_1= r_2= R, \, r_3=\infty \,; \rightarrow r = R/4. $$

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