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How can I find the diameter of a circle that's been rolled up to a wall when the circle is touching a rectangle in the corner of the wall with height $8$ and width $5$?

Here's a picture of what I mean:

enter image description here

I tried drawing a right triangles in several different areas, but none of them helped me out. I'm pretty sure that's the right approach, but I still can't figure out this problem.

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    $\begingroup$ It seems that with the information you have given, you could draw many possible circles that fit the conditions (like the one you have drawn and any circle larger than it). There must be an additional piece of information to answer your question. Perhaps is the corner touching point the left-most point of the circle (and so its radius is 8?) $\endgroup$ – JimN Sep 8 '20 at 5:32
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    $\begingroup$ It says, "The circle is rolled until it's flush with the wall". My image is inaccurate. I'm really sorry. So it's touching the wall and the corner. I just updated the post. Also, I know that it just barely touches the box pressed into the corner of the wall. $\endgroup$ – user822241 Sep 8 '20 at 5:43
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    $\begingroup$ So are you able to answer the question now that you have the correct interpretation? $\endgroup$ – JimN Sep 8 '20 at 5:52
  • $\begingroup$ I've tried for another 10 minutes, and I'm still not able to. Let $R$ be the radius. Then if we draw a line from the top right of the rectangle to the center (length $R$) and drop straight down (length $R + 5$), we get an unknown side-length equal to $x$. I've also tried another way in which I get side-lengths $R, R + 8$ and $y$. I can't figure much else out from here. $\endgroup$ – user822241 Sep 8 '20 at 6:12
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    $\begingroup$ The midpoint of the circle is $(R,R)$ where $R$ is the radius. Moreover $(5-R)^2+(8-R)^2=R^2$. $\endgroup$ – Jens Schwaiger Sep 8 '20 at 7:10
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Let the tangents be the coordinate axes and hence the intersection point be the origin.

You get the equation of circle as $$(x-r)^2+(y-r)^2=r^2$$ This circle passes through $(5,8)$. You can find $r$ by these information.


Alternate method:

Use Pythagoras theorem.

enter image description here

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  • $\begingroup$ By plugging in $(x, y) = (5, 8)$, I end up getting $r = 13 \pm 4\sqrt{5}$. Both solutions are positive. How do I know which is right? $\endgroup$ – user822241 Sep 8 '20 at 7:15
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    $\begingroup$ @Wan, $r$ should be greater than $5$ because $r$ is $x$-coordinate of center of the circle. $\endgroup$ – SarGe Sep 8 '20 at 7:20
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I think you might be misinterpreting your question. As it is stated, there is no unique solution since there are many possible circles that satisfy your configuration. But since you write that "the circle that's been rolled up to a wall" I think the picture should look more like:

enter image description here

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  • $\begingroup$ I think if he is tangent to both walls and passes trough that vertice of the rectangle then it's radius is defined $\endgroup$ – hellofriends Sep 8 '20 at 5:41
  • $\begingroup$ Yes, your image is correct. I'm really sorry. It says, "The circle is rolled until it's flush with the wall". My image is not accurate. Also, I know that it just barely touches the box pressed into the corner of the wall. $\endgroup$ – user822241 Sep 8 '20 at 5:43
  • $\begingroup$ This is the correct interpretation. $\endgroup$ – Gray_Rhino Sep 8 '20 at 5:45
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Geometric solution.

enter image description here

Start with the rectangle $ABCD$, $|AC|=|BD|=5$, $|AD|=|BC|=8$. Its diagonal $AB$ makes the right-angled $\triangle ABC$ with $|AB|=\sqrt{89}$. Construct the inscribed circle and its center $I$.

For the reference, its radius is

\begin{align} r&=\tfrac12\,(|AC|+|BC|-|AB|) =\tfrac{13}2-\tfrac12\,\sqrt{89} \tag{1}\label{1} , \end{align}
and the center is $I(r,\, r)$, assuming that $C(0,0)$ is the origin.

Next, find the point $E$ as the lower intersection of the line $CD$ with the incircle.

The center of the sought circle is found at the intersection of the line $CI$ (the bisector of the $\angle BCA$) and the line through the point $D$ parallel to the line $EI$.

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