I'm trying to figure out a couple things. The main question I have is how to find the area of a circle inscribed inside a quarter circle with a radius of x. The secondary question to that is if the radius of the inner circle drawn to touch the tangent lines of the sides of the quarter circle bisects the sides of the quarter circle. I hope that makes sense...

  • $\begingroup$ <irrelevant> I asked myself this question a few years ago and wasn't able to figure it out ever since - seeing it here is quite the pleasant surprise. </irrelevant> $\endgroup$ – theage Jan 2 '15 at 22:17
  • $\begingroup$ There's no like button here, so I'll just comment to say it. I'd wager there was a stick that became dislodged from your brain today. $\endgroup$ – Joseph Jan 2 '15 at 22:36

A figure helps a lot. The three segments labeled $r$ are all radii of the small circle. The diagonal of the square is $r\sqrt 2$, so $r(1+\sqrt 2)=x$ No, the tangency points do not bisect the radii of the big circle, they cut them in ratio $1: \sqrt 2$

enter image description here

  • $\begingroup$ A figure is good for sure. I'm new to this forum, so I don't really know how to add pictures. So, where does the "1" come from in the "r(1+sqrt(2))" $\endgroup$ – Joseph Jan 2 '15 at 22:22
  • $\begingroup$ @Joseph, the $r(1 + \sqrt(2))$ comes from the $r \sqrt{2}$ needed to cross the square and the additional $r$ needed to get to the edge of the circle (going perfectly NE/SW) $\endgroup$ – Charles Baker Jan 2 '15 at 22:24
  • $\begingroup$ x is equal to the radius of the small circle plus the diagonal of the square. $\endgroup$ – Brady Gilg Jan 2 '15 at 22:24
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    $\begingroup$ The radius of the quarter-circle, $x$, is equal to the diagonal of the square plus the radius of the small circle, which is $r\sqrt{2}+r$. This factors to $r(1+\sqrt{2})$. $\endgroup$ – KSmarts Jan 2 '15 at 22:26
  • $\begingroup$ Got it, thanks. Simple factoring. Silly me, lol. $\endgroup$ – Joseph Jan 2 '15 at 22:31

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