Have a circle with a radius of 1. Split the circle into 4 equal quadrants. Inside 1, draw a circle that touches both radius and the circle arc. How do I calculate the radius of smaller circle ?

Cheers

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- Nov 3rd 2012, 12:26 AMArchiecrabCircle problem
Have a circle with a radius of 1. Split the circle into 4 equal quadrants. Inside 1, draw a circle that touches both radius and the circle arc. How do I calculate the radius of smaller circle ?

Cheers - Nov 3rd 2012, 01:38 AMearbothRe: Circle problem
1. Draw a sketch! (see attachment)

2. You've got 2 right triangles. The larger one has the hypotenuse

$\displaystyle 2r+x=1$ ....................**[A]**

3. Use proportions:

$\displaystyle \frac{r}{\frac12 \sqrt{2}}=\frac{r+x}1$ .................... [B]

4. From [A] you get: $\displaystyle x=1-2r$

Thus your proportion becomes:

$\displaystyle \frac{r}{\frac12 \sqrt{2}}=\frac{r+1-2r}1$

$\displaystyle \frac{r}{\frac12 \sqrt{2}}=1-r$

$\displaystyle r+r \cdot \frac12 \sqrt{2}=\frac12 \sqrt{2}~\implies~\boxed{r=\frac{\frac12 \sqrt{2}}{1+\frac12 \sqrt{2}}}$ - Nov 3rd 2012, 02:06 AMMarkFLRe: Circle problem
I used an identical drawing and determined the point where the two circles touch must have equal coordinates, and so we may determine:

$\displaystyle x=y=\frac{1}{\sqrt{2}}$

And thus we may state:

$\displaystyle 2(x-r)^2=r^2$

After simplification we find:

$\displaystyle r=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$ - Nov 3rd 2012, 06:34 AMHallsofIvyRe: Circle problem
And, remarkably, those two answers are the same!

Starting with earboth's result, $\displaystyle \frac{\frac{1}{2}\sqrt{2}}{1+ \frac{1}{2}\sqrt{2}}$, multiply both numerator and denominator by 2 to get $\displaystyle \frac{\sqrt{2}}{2+ \sqrt{2}}$. Now, "rationalize" the denominator by multiplying both numerator and denominator by $\displaystyle 2- \sqrt{2}$:

$\displaystyle \frac{\sqrt{2}}{2+ \sqrt{2}}\frac{2- \sqrt{2}}{2- \sqrt{2}}= \frac{2\sqrt{2}- 2}{4- 2}= \frac{2(\sqrt{2}- 1)}{2}= \sqrt{2}- 1$