Radius of a circle touching a rectangle both of which are inside a square

Mooncrater 09/13/2018. 5 answers, 1.020 views
geometry circle rectangles

Given this configuration :

enter image description here

We're given that the rectangle is of the dimensions 20 cm by 10 cm, and we have to find the radius of the circle.

If we somehow know the distance between the circle and the corner of the square then we can easily find the radius. (It's equal to $ \sqrt{2}\times R-R$)

I really can't understand how to solve it. Any help appreciated.

5 Answers


Seyed 09/14/2018.

It is just using the pythagorean theorem:
$a=10$ $cm$
$b=20$ $cm$
$(r-a)^2+(r-b)^2=r^2$
$(r-10)^2+(r-20)^2=r^2$
$r^2+100-20r+r^2+400-40r=r^2$
$r^2-60r+500=0$
$r=50$ $cm$
$r=10$ $cm$
The $r=50$ $cm$ is the acceptable answer.

enter image description here


Doug M 09/13/2018.

Place the center of the cricle at $O.$

Let the radius be $R$

The corner of the square is $(R,R)$ I have made a small alteration to the picture to create fewer negative numbers.

Offsetting by the rectange, the corer of the rectangle is $(R-20, R-10)$

And the distance from this point equals the $R.$

That should put you on your way to the solution.


Phil H 09/13/2018.

$(x, y) = (R-20, R-10)$ as a point on the circle $y = \sqrt{R^2 - x^2}$

$R - 10 = \sqrt{R^2 - (R-20)^2}$

$(R- 10)^2 = R^2 - (R-20)^2$

$R^2 - 20R + 100 = R^2 - (R^2 - 40R + 400)$

$R^2 - 60R + 500 = 0$

$(R - 50)(R-10) = 0$

$R = 50$ is the only sensible option.


Danijel 09/14/2018.

The point where rectangle touches the circle is $|R-a|$ and $|R-b|$ away from $x$ and $y$ axes, where $a$ and $b$ are lengths of sides of the rectangle and $R$ is the radius of the circle.

This leads to the equation $$(R-a)^2 + (R-b)^2 = R^2,$$ which has solutions $$R_{1,2} = a+b\pm\sqrt{2ab}.$$

One solution corresponds to a bigger rectangle (compared to the circle), one touching the circle on the other side, which is not the case here. Smaller rectangle compared to the circle means that the circle is bigger if the rectangle is kept fixed, so the correct radius is $$R = a+b+\sqrt{2ab}.$$

Plugging in $a=10$ and $b=20$ gives $R=50$.


R. Romero 09/13/2018.

You can use trig to get the same answer as those above.

Draw three lines: One from the center of the circle to the corner shared by the square and the rectangle. Then draw a line from the center of the circle to the corner nearest to the center of the circle. Draw a final line being the diagonal connecting the previously mentioned corners.

We know the length of the third line by the pythagorean theorem. If we call the side length of the square L, the length of the shorter of the two remaining lines is L/2. The length of the longer, L/sqrt(2).

Find the angle that the diagonal makes with the longer of the drawn lines allows you to apply the cosine rule.

The longer line meets the square's corner at a 45 degree angle with respect to either side. Then angle the diagonal makes with the left side of the square has a tangent of 2.

Apply the cosine rule then solve the resulting quadratic and you get two possible answers, only one of which is plausible.


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