What is the radius of the red circle?

Question

The blue circle has radius $2$ and the green circle has radius $1$. $AB$ is a common tangent and all three circles touch each other. Find the radius of the red circle.

Answer 1

Let $C_1$ be the circle of radius $1$ and denoted by $O_1$ its center. Similarly, call $C_2$ be the circle of radius $2$ and denote by $O_2$ its center. The distance $\overline{O_1O_2}$ is easy to see that it's equal to $3$. Let $C$ the red circle and call its center $O$ and set $x$ its radius. We would like to compute $x$. Notice that the length of $\overline{OO_1}$ is $1+x$, similarly the length of $\overline{OO_2}$ is $2+x$. Let $F_1$ (resp. $F_2$) be the feet of the unique straight line passing through $O_1$ (resp. $=_2$) and orthogonal to $AB$. Call $y$ the distance between $F_1$ and $F_2$. Using the Pythagorean theorem three times, you can see that $$ 2\sqrt{2}=2\sqrt{2x}+2\sqrt{x}$$ Indeed, let $M$ be the feet of the unique line passing through $O$ and orthogonal to $AB$. Call $y_1$ the distance between $M$ and $F_1$ and call $y_2$ the distance between $M$ and $F_2$. Finally call $y$ the distance between $F_1$ and $F_2$. By using the Pythagorean theorem you can deduce that $y=2\sqrt{2}$, $y_1=2\sqrt{x}$ and $y_2=2\sqrt{2x}$, then use the fact that $y=y_1+y_2$. Simplyfing the equation above you get

$$\sqrt{2}=\sqrt{2x}+\sqrt{x}$$ iff $$\sqrt{2}=\sqrt{x}(\sqrt{2}+1)$$ iff $$2=x(\sqrt{2}+1)^2$$ iff $$x=\frac{2}{(\sqrt{2}+1)^2}$$

EDIT: After Roost1513's comment about the presence of the radical at the denominator, I decide to simplify further the value of $x$ in order to have a nicer value of $x$, despite the answer was complete yet. Precisely

$$x=\frac{2}{(\sqrt{2}+1)^2}=\frac{2(\sqrt{2}-1)^2}{(\sqrt{2}+1)^2(\sqrt{2}-1)^2}=6-4\sqrt{2}$$

Answer 2

I think I could explain this a bit clearer.

Let the center of the red circle be O0, let the center of the green circle be O1, and let the center of the blue circle be O2. Let where the blue circle touches the line AB be C, let where the green circle touches AB be D, and let where the red circle touches AB be E. Let the radius of the red circle be X.

We know this:

O1 to O2 = 3

O1 to D = 1

O2 to C = 2

And we also know this, but it will become more relevant later:

O0 to E = X

O1 to O0 = 1 + X

O2 to O0 = 2 + X

We should attempt to find the length of C to D, in order to understand the entire shape clearer. In order to do that, as we know that the angle of O2 to C to D is 90 degrees and the angle of O1 to D to C is also 90 degrees, we can imagine a triangle from O2 to O1 to a point half the distance between O2 to C, of which we will call F. The sides of this triangle would be:

O2 to O1 = 2 + 1 = 3

O2 to F = O2 - O1 = 2 - 1 = 1

F to O1 = CD.

Using the pythagorean theorem, we know that $3^2 = 1^1 + CD^2$, or $9 - 1 = CD^2$. Therefore, CD equals the square root of 8, or $2\sqrt{2}$.

The following may get a little confusing. In order to form two triangles, we need to shift the entire AB line up to the center of O0, or the entire length of X. For the purposes of this demonstration, the two important impacted points will now be called C1 and D1.

The first triangle formed is O2 to C1 to O0. The lengths would be:

O2 to C1 = 2 - X

O2 to O0 = 2 + X

C1 to O0 = Y

The second triangle formed is O1 to D1 to O0. The lengths would be:

O1 to D1 = 1 - X

O1 to O0 = 1 + X

D1 to O0 = Z

We also know that, as we know Y + Z = C1 to O0 + D1 to O0 = C1 to D1 = CD = $2\sqrt{2}$, Y + Z = $2\sqrt{2}$.

We will now solve for both Y and Z using a slightly modified version of the pythagorean theorem.

Y:

$$(2+X)^2 - (2-X)^2 = Y^2$$

$$(4+4X+X^2) - (4-4X+X^2) = Y^2$$

$$8X = Y^2$$

$$2\sqrt{2X} = Y$$

Z:

$$(1+X)^2 - (1-X)^2 = Z^2$$

$$(1+2X+X^2) - (1-2X+X^2) = Z^2$$

$$4X = Z^2$$

$$2\sqrt{X} = Z$$

Aha! We're nearly there! As we know Y + Z = $2\sqrt{2}$, we can also write...

$$2\sqrt{2X} + 2\sqrt{X} = 2\sqrt{2}$$

$$\sqrt{2X} + \sqrt{X} = \sqrt{2}$$

$$\sqrt{X}(\sqrt{2}+1) = \sqrt{2}$$

$$X(\sqrt{2}+1)^2 = 2$$

$$X = \frac{2}{(\sqrt{2}+1)^2}$$

$$X = \frac{2}{(2 + 2\sqrt{2} + 1)}$$

$$X = \frac{2}{(3 + 2\sqrt{2})}$$

$$X = \frac{2(3 - 2\sqrt{2})}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})}$$

$$X = \frac{2(3 - 2\sqrt{2})}{(9-4*2)}$$

$$X = 6 - 4\sqrt{2}$$

Answer 3

Let the radius of the red circle be $r cm$.

By tangent property, we have $$CD-DE \textrm{ and } EF=FG \Rightarrow \frac{r}{\tan \alpha}=2 \tan \alpha \textrm{ and } \frac{r}{\tan \theta}=\tan \theta $$

Eliminating $r$ yields $$\tan \theta =\sqrt{2} \tan \alpha \tag*{(*)} $$

Applying Pythagoras’ Theorem on $\triangle HKL$ yields

$$ \sqrt{3^{2}-1^{2}}=2(2 \tan \alpha+\tan \theta). $$

Putting (*) gives $$ \begin{aligned} \sqrt{8} &=2(2 \tan \alpha+\sqrt{2} \tan \alpha) \\ \tan \alpha &=\frac{\sqrt{2}}{2+\sqrt{2}}\\&=\frac{2-\sqrt{2}}{\sqrt{2}} \\ r&=2\tan^2 \alpha \\&=2\left(\frac{2-\sqrt{2}}{\sqrt{2}}\right)^{2} \\ &=6-4 \sqrt{2} \end{aligned} $$

In general, let the radii of the blue, green and red circles be $a, b$ and $r$ respectively. performing the same computation yields

$$ \boxed{r=\frac{a b^{2}}{1+a b+2 \sqrt{a b}}} $$

Answer 4

You have three circles, of curvatures $k_1 = \frac{1}{2}, k_2 = 1 , k_3 = 0 $

So by Descartes kissing circles theorem, the curvature of the small circle is

$\begin{equation} \begin{split} k_4 &= k_1 + k_2 + k_3 + 2 \sqrt{k_1 k_2 + k_1 k_3 + k_2 k_3 } \\ &= \dfrac{3}{2} + 2 \sqrt{ \dfrac{1}{2}}\\ &= \dfrac{ 3 \sqrt{2} + 4 }{2 \sqrt{2}}\\ \end{split} \end{equation} $

Hence, the radius is

$r_4 = \dfrac{1}{k_4} = \dfrac{ 2 \sqrt{2} }{3 \sqrt{2} + 4} = \dfrac{12 - 8\sqrt{2}}{2} = 6 - 4 \sqrt{2} $

Answer 5

In general, the radius $r$ of a circle inscribed by any two touching circles of radii $a$ & $b$ and their common tangent, is given by Generalized Formula as follows $$\boxed{r=\frac{ab}{(\sqrt a+\sqrt b)^2}}$$ Now, substituting the values of radii $a=2$ of blue circle and $b=1$ of green circle, the radius $r$ of red circle is calculated as follows $$r=\frac{2\cdot1}{(\sqrt 2+\sqrt 1)^2}$$ $$=\frac{2}{3+2\sqrt2}$$ $$=\frac{2(3-2\sqrt2)}{(3+2\sqrt2)(3-2\sqrt2)}$$ $$=2(3-2\sqrt2)$$