Circle problem

**Archiecrab** · November 3rd 2012, 12:26 AM

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

**earboth** · November 3rd 2012, 01:38 AM

Originally Posted by Archiecrab

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

1. Draw a sketch! (see attachment)

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

$2r+x=1$ .................... [A]

3. Use proportions:

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

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

Thus your proportion becomes:

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

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

$r+r \cdot \frac12 \sqrt{2}=\frac12 \sqrt{2}~\implies~\boxed{r=\frac{\frac12 \sqrt{2}}{1+\frac12 \sqrt{2}}}$

**MarkFL** · November 3rd 2012, 02:06 AM

I used an identical drawing and determined the point where the two circles touch must have equal coordinates, and so we may determine:

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

And thus we may state:

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

After simplification we find:

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

**HallsofIvy** · November 3rd 2012, 06:34 AM

And, remarkably, those two answers are the same!
Starting with earboth's result, $\frac{\frac{1}{2}\sqrt{2}}{1+ \frac{1}{2}\sqrt{2}}$ , multiply both numerator and denominator by 2 to get $\frac{\sqrt{2}}{2+ \sqrt{2}}$ . Now, "rationalize" the denominator by multiplying both numerator and denominator by $2- \sqrt{2}$ :
$\frac{\sqrt{2}}{2+ \sqrt{2}}\frac{2- \sqrt{2}}{2- \sqrt{2}}= \frac{2\sqrt{2}- 2}{4- 2}= \frac{2(\sqrt{2}- 1)}{2}= \sqrt{2}- 1$

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