Find radius of a circle using stewart theorem

Solution 1:

Just as @Riverboat suggested.

Consider $\triangle{C_1C_2C_3}$

Clearly, $C_1C_2=5$

Let the radius of circle with centre $C_3$ be $r$

So, $C_1C_3=r+3$ and $C_2C_3=r+2$

With the base, $b=C_1C_2=5$ and height, $h=5-r$, we get

$$[\triangle{C_1C_2C_3}]=\frac{1}{2} \cdot 5 \cdot (5-r)\tag{1}$$

Using Heron's Formula, where

$$a=C_1C_2=5, b=C_1C_3=r+3 \text{ and }c=C_2C_3=r+2$$

$$s=\frac{a+b+c}{2}=r+5$$

We get

$$[\triangle{C_1C_2C_3}]=\sqrt{(5+r)\cdot 2 \cdot 3 \cdot r}\tag{2}$$

Using $(1)$ and $(2)$, we get

$$6(r^2+5r)=\frac{25}{4} \cdot (5-r)^2 \Longleftrightarrow r^2-370r+625=0$$

Solving, we get $r=5(37\pm8\sqrt{21})$

But since $C_3$ touches $C$ internally, $r \leq 5$

Thus, $\color{red}{r=5(37-8\sqrt{21})}$

An alternative could be Descartes' Circle Theorem

Using this, we get

$$(k_1+k_2+k_3+k_4)^2=2(k_1^2+k_2^2+k_3^2+k_4^2)$$

Here, $k_i$ represents the curvature and is given by $$k_i=\pm \dfrac{1}{r_i}$$ where

• $+$ sign is given to a circle externally tangent to other circles.
• $-$ sign is given to a circle internally tangent to other circles.

So, $$k_1=-\frac{1}{5}, k_2=\frac{1}{3} \text{ and } k_3=\frac{1}{2}$$

Solving the quadratic in $k_4$, we get two values of which the positive one is the required curvature. Let this value be $m$.

Thus, the radius of the circle is $r_4=\dfrac{1}{m}$