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I have one circle of $11.63$ inch bricks that is $21.25$ ft in radius (green in the picture). I already purchased the bricks for this circle already.

I need to add another circle to the side (marked in red) of bricks. The size is scaled by the golden ratio of $1.618$, or $13.133$ ft in radius. The center of this red circle falls directly on the edge of the green circle.

I need to purchase bricks to form the second (red) circle, but I don't need bricks to go inside the green circle.

How do I find the circumference of circle B, minus the part intersected with circle A, so I can calculate how many bricks to buy?

cosmo5
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Village
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2 Answers2

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The $36^{\circ}-72^{\circ}-72^{\circ}$ $\triangle ABC$ is called the golden triangle as its sides are $a/\phi,a,a$.

Desired red circumference is $$2\pi\cdot \frac{a}{\phi} \cdot\left(1-\frac{2\cdot72}{360}\right)$$ $$=2\pi\cdot \frac{a}{\phi}\cdot \frac{3}{5}$$ $$=\frac{6\pi}{5}\cdot \frac{a}{\phi} $$

where $a=21.25$

cosmo5
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One possibility is to think about the intersection of the two circles. Assuming that the red circle is centered at $(0,0)$, its ''northeast'' quadrant is described by $y=\sqrt{r^2-x^2}$ (where $r$ is its radius). The green circle is centered at $(R,0)$, $R$ being its radius, and its ''northwest'' quadrant is described by $y=\sqrt{R^2-(x-R)^2}$. If you solve

$$\sqrt{r^2-x^2}=\sqrt{R^2-(x-R)^2}$$

you obtain the points at which the circles intersect: $\left(\frac{r^2}{2R},r\sqrt{1-\frac{r^2}{4R^2}}\right)$ and, by symmetry, $\left(\frac{r^2}{2R},-r\sqrt{1-\frac{r^2}{4R^2}}\right)$. These two points define an arc that spans $2\arctan\frac{r\sqrt{1-\frac{r^2}{4R^2}}}{\frac{r^2}{2R}}$ of the $360ยบ$.

Patricio
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