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A circle C of radius 5 cm and two circles C1 and C2 of radius 3,2 respectively . C1C2 touch each other externally and both touch C internally . A circle C3 touch C1,C2 externally and touch C internally of radius r . We have to find radius r .

, i tried it stewart theorm but in that not able to get cevian length

Ethan Bolker
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Koolman
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  • Aren't $C$ and $C_3$ the same in this case. Anyways, please include a diagram to make this question clear and show us where you are stuck. – Roby5 Jul 26 '16 at 11:20
  • @Roby5 no they are not same – Koolman Jul 26 '16 at 11:22
  • Stewart's theorem is enough. Have you tried triangleC1C2C3 (each represents center of their respective circle) with cevian CC3=5-radius of C3? – Ahmbak Jul 26 '16 at 11:25
  • @Riverboat but what is the length of cevian – Koolman Jul 26 '16 at 11:26
  • @koolman Let r be the radius of C3. Consider triangleC1C2C3 which has side lengths 5,2+r,3+r. The cevian we talk about is CC3 with length = 5-r(since C,C3 is tangent internally). The remaining part is just solving the equation. – Ahmbak Jul 26 '16 at 11:31
  • @Riverboat is CC3 is tangent to C1 &C2 – Koolman Jul 26 '16 at 12:00
  • @koolman No. We can compute CC1=5-3 and CC2=5-2. Now you just plug these values into Stewart's theorem. I got 30/19 as an answer. – Ahmbak Jul 26 '16 at 13:50
  • @Riverboat can you give me the answer with figure – Koolman Jul 26 '16 at 13:52

1 Answers1

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 $r_4=\dfrac{1}{m}$

Roby5
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