# Two tangent circles are inscribed in a semicircle, one touching the diameter's midpoint; find the radius of the smaller circle I am unable to upload the image of my trials.

I assumed the radius of small circle is $$x,$$ horizontal distance between the centers of two circles is $$y.$$

I have joined the centers of the two circles and the length is $$(5+x).$$

I have drawn a vertical line from the center of the bigger circle to the center of the semi circle.

I have also drawn a horizontal line from the center of the small circle to the above line.

Then, by applying Pythagoras theorem, I get
$$(x+5)^2=(5-x)^2 + y^2.$$ I need one more equation to solve for $$x.$$

Intuitively I wonder if the radius of the small circle could be half of the big circle. I carefully constructed it and got $$2.5 cm$$ as the radius but I am not sure.

• Hey Ashwini. You should be able to type out your working-out in your trials. People do appreciate when you show effort in trying to solve the question and it helps viewers solve the answers for you. – MBorg Nov 27 '18 at 9:39
• Okay. Let me try – Ashwini Nov 27 '18 at 10:00
• Radius of small circle = x Horizontal distance between two circles= y. I have joined centers of the two circles and the length is (5+x). I have drawn a vertical line from the center of big circle to center of semi circle. I have also drawn horizontal line from center of small circle to the above line. Then ,by applying Pythagoras theorem, I get – Ashwini Nov 27 '18 at 10:06
• (x+5)^2=(5-x)^2 + y^2. I need one more equation to solve for x.. Intuitively I wonder if the radius of the small circle could be half of the big circle. I carefully constructed it and got 2.5 cm as the radius but I am not sure. – Ashwini Nov 27 '18 at 10:10
• You should edit your working-out into your question, to improve its quality and for easier-viewing :) – MBorg Nov 27 '18 at 23:35

In addition to using Pythagorean theorem, one can use circle inversion to figure out the radius of the small circle (let's call it $$r$$). Let $$AB$$ be the base of the semicircle. Let $$O$$ be its midpoint. Let $$C$$ be the contact point between the small circle (blue) with the circular arc $$AB$$. Draw the line $$OC$$ and let $$D$$ be its other intersection with the small circle. $$CD$$ will be a diameter of the small circle.

Perform a circle inversion with respect to the circle centered at $$O$$ with radius $$10$$. The line $$AB$$ get mapped to itself. The big circle (green) get mapped to a line (green, dashed) parallel to $$AB$$ and at a distance $$10$$ from it. The small circle get mapped to a circle (blue, dashed) sandwiched between these two lines. So its diameter will be $$10$$. Let $$D'$$ be the image of $$D$$ under circle inversion. $$CD'$$ will be a diameter of the image of the small circle. We have

$$|CD'| = 10 \implies |OD'| = |OC|+|CD'| = 10 + 10 = 2|OC|$$ Circle invert $$OD'$$ back to $$OD$$, we find \begin{align}|OD| = \frac12|OC| &\implies |CD| = |OC| - |OD| = \frac12|OC|\\ &\implies r = \frac12|CD| = \frac14|OC| = \frac52 \end{align} The radius we seek is $$\frac52$$. One half of that of the big circle and a quarter of that of the semicircle.

• This means the radius of the small circle will be half the radius of big circle for any given radius. – Ashwini Nov 28 '18 at 13:34
• @Ashwini yes, under the assumption the radius of big circle is one half of that of the semi-circle, – achille hui Nov 28 '18 at 13:48

Let the radius of smaller circle be $$\displaystyle r$$ and x-coordinate of its center$$\displaystyle ( C)$$ is $$\displaystyle a$$. As the circle is touching x-axis, so the ordinate of center of cicle is equal to radius of the circle,i.e., $$\displaystyle r$$. Let the point of touching of semicircle and smaller circle be $$\displaystyle P_{1}$$ $$\displaystyle ( x_{1}\displaystyle ,y_{1})$$. As the semicircle and smaller circle touch each other so $$\displaystyle P_{1}$$, $$\displaystyle C,Origin$$ are collinear. $$\begin{gather*} \therefore \dfrac{r}{a} =\dfrac{y_{1}}{x_{1}} \ \ \ \ \ \ \ \ \ \ \ ( 1)\\ \end{gather*}$$

And the distance between $$\displaystyle C$$ and center of bigger circle is $$\displaystyle r+5$$ $$\begin{gather*} ( r-5)^{2} +a^{2} =( r+5)^{2}\\ or\ \ 20r=a^{2} \ \ \ \ \ \ \ \ \ \ ( 2) \end{gather*}$$ Also the point $$\displaystyle P_{1}$$ satisfies both the semicircle and the smaller circle. $$\begin{gather*} \therefore x^{2}_{1} +y^{2}_{1} =100\ \ \ \ \ \ \ ( 3)\\ And\ ( x_{1} -a)^{2} +( y_{1} -r)^{2} =r^{2}\\ or\ x^{2}_{1} +y^{2}_{1} +a^{2} -2ax_{1} -2ry_{1} =0\\ or\ 100+a^{2} -2ax_{1} -2ry_{1} =0\ \ \ \ \ \ \ \ \ ( 4) \end{gather*}$$ Solving these four equations, we get $$\displaystyle r=2.5\ units$$

• Yes. That means we should use coordinate geometry. And the radius happens to be half of the big circle. This maybe a symmetrical property of circle. – Ashwini Nov 27 '18 at 10:19
• @Ashwini I don't think so. It is just a coincidence that its half of radius of bigger circle – Dikshit Gautam Nov 27 '18 at 12:11
• @DikshitGautam this cannot be a coincidence. – user376343 Nov 29 '18 at 20:34

Let $$R$$ be the (known) radius of the large inscribed circle, $$r$$ the radius of the small inscribed circle, and $$(x,r)$$ the center of this small circle. Then one has the two equations $$x^2+(R-r)^2=(R+r)^2,\qquad\sqrt{x^2+r^2}+ r=2R$$ in the two unknowns $$r$$ and $$x$$.

• Yes I understand. there are 2 unknowns here r and x. how can I proceed further to get r? Or is it that there is insufficient information to solve this problem? – Ashwini Nov 27 '18 at 10:14

With hindsight - never any use, of course! - one can see that such a configuration must exist, because there exists an isosceles triangle whose sides are in the ratio of $$3:2$$ (whose height and angles one needn't know): 