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Find radius of a circle using stewart theorem

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0

$begingroup$

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

euclidean-geometry

share|cite|improve this question

edited Jul 26 '16 at 14:40

Ethan Bolker

45.7k553120

asked Jul 26 '16 at 11:08

KoolmanKoolman

1,471824

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Dragonemperor42
  Jul 26 '16 at 11:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Roby5 no they are not same
  $endgroup$
  – Koolman
  Jul 26 '16 at 11:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Stewart's theorem is enough. Have you tried triangleC1C2C3 (each represents center of their respective circle) with cevian CC3=5-radius of C3?
  $endgroup$
  – Ahmbak
  Jul 26 '16 at 11:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Riverboat but what is the length of cevian
  $endgroup$
  – Koolman
  Jul 26 '16 at 11:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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.
  $endgroup$
  – Ahmbak
  Jul 26 '16 at 11:31
  

  
  
  
  

 | 
show 3 more comments

0

$begingroup$

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

euclidean-geometry

share|cite|improve this question

edited Jul 26 '16 at 14:40

Ethan Bolker

45.7k553120

asked Jul 26 '16 at 11:08

KoolmanKoolman

1,471824

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Dragonemperor42
  Jul 26 '16 at 11:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Roby5 no they are not same
  $endgroup$
  – Koolman
  Jul 26 '16 at 11:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Stewart's theorem is enough. Have you tried triangleC1C2C3 (each represents center of their respective circle) with cevian CC3=5-radius of C3?
  $endgroup$
  – Ahmbak
  Jul 26 '16 at 11:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Riverboat but what is the length of cevian
  $endgroup$
  – Koolman
  Jul 26 '16 at 11:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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.
  $endgroup$
  – Ahmbak
  Jul 26 '16 at 11:31
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

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

euclidean-geometry

share|cite|improve this question

edited Jul 26 '16 at 14:40

Ethan Bolker

45.7k553120

asked Jul 26 '16 at 11:08

KoolmanKoolman

1,471824

$endgroup$

share|cite|improve this question

edited Jul 26 '16 at 14:40

Ethan Bolker

45.7k553120

asked Jul 26 '16 at 11:08

KoolmanKoolman

1,471824

share|cite|improve this question

edited Jul 26 '16 at 14:40

Ethan Bolker

45.7k553120

asked Jul 26 '16 at 11:08

KoolmanKoolman

1,471824

edited Jul 26 '16 at 14:40

Ethan Bolker

45.7k553120

edited Jul 26 '16 at 14:40

Ethan Bolker

45.7k553120

asked Jul 26 '16 at 11:08

KoolmanKoolman

1,471824

asked Jul 26 '16 at 11:08

KoolmanKoolman

1,471824

1 Answer
1

active

oldest

votes

0

$begingroup$

Just as @Riverboat suggested.

Consider $triangleC_1C_2C_3$

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

$$[triangleC_1C_2C_3]=frac12 cdot 5 cdot (5-r)tag1$$

Using Heron's Formula, where

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

$$s=fraca+b+c2=r+5$$

We get

$$[triangleC_1C_2C_3]=sqrt(5+r)cdot 2 cdot 3 cdot rtag2$$

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

$$6(r^2+5r)=frac254 cdot (5-r)^2 Longleftrightarrow r^2-370r+625=0$$

Solving, we get $r=5(37pm8sqrt21)$

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

Thus, $colorredr=5(37-8sqrt21)$

---

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 $$k_i=pm dfrac1r_i$$ where

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


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

So, $$k_1=-frac15, k_2=frac13 text and k_3=frac12$$

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 $r_4=dfrac1m$

share|cite|improve this answer

edited Jul 26 '16 at 14:54

answered Jul 26 '16 at 11:53

Dragonemperor42Dragonemperor42

4,0431925

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not what I meant though. How do you know CC3 is the height of triangle C1C2C3? Anyway, thanks for sharing Descartes's theorem :).
  $endgroup$
  – Ahmbak
  Jul 26 '16 at 13:54
  

  
  
  
  

add a comment | 

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0

$begingroup$

Just as @Riverboat suggested.

Consider $triangleC_1C_2C_3$

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

$$[triangleC_1C_2C_3]=frac12 cdot 5 cdot (5-r)tag1$$

Using Heron's Formula, where

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

$$s=fraca+b+c2=r+5$$

We get

$$[triangleC_1C_2C_3]=sqrt(5+r)cdot 2 cdot 3 cdot rtag2$$

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

$$6(r^2+5r)=frac254 cdot (5-r)^2 Longleftrightarrow r^2-370r+625=0$$

Solving, we get $r=5(37pm8sqrt21)$

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

Thus, $colorredr=5(37-8sqrt21)$

---

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 $$k_i=pm dfrac1r_i$$ where

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


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

So, $$k_1=-frac15, k_2=frac13 text and k_3=frac12$$

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 $r_4=dfrac1m$

share|cite|improve this answer

edited Jul 26 '16 at 14:54

answered Jul 26 '16 at 11:53

Dragonemperor42Dragonemperor42

4,0431925

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not what I meant though. How do you know CC3 is the height of triangle C1C2C3? Anyway, thanks for sharing Descartes's theorem :).
  $endgroup$
  – Ahmbak
  Jul 26 '16 at 13:54
  

  
  
  
  

add a comment | 

0

$begingroup$

Just as @Riverboat suggested.

Consider $triangleC_1C_2C_3$

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

$$[triangleC_1C_2C_3]=frac12 cdot 5 cdot (5-r)tag1$$

Using Heron's Formula, where

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

$$s=fraca+b+c2=r+5$$

We get

$$[triangleC_1C_2C_3]=sqrt(5+r)cdot 2 cdot 3 cdot rtag2$$

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

$$6(r^2+5r)=frac254 cdot (5-r)^2 Longleftrightarrow r^2-370r+625=0$$

Solving, we get $r=5(37pm8sqrt21)$

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

Thus, $colorredr=5(37-8sqrt21)$

---

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 $$k_i=pm dfrac1r_i$$ where

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


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

So, $$k_1=-frac15, k_2=frac13 text and k_3=frac12$$

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 $r_4=dfrac1m$

share|cite|improve this answer

edited Jul 26 '16 at 14:54

answered Jul 26 '16 at 11:53

Dragonemperor42Dragonemperor42

4,0431925

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not what I meant though. How do you know CC3 is the height of triangle C1C2C3? Anyway, thanks for sharing Descartes's theorem :).
  $endgroup$
  – Ahmbak
  Jul 26 '16 at 13:54
  

  
  
  
  

add a comment | 

$begingroup$

Just as @Riverboat suggested.

Consider $triangleC_1C_2C_3$

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

$$[triangleC_1C_2C_3]=frac12 cdot 5 cdot (5-r)tag1$$

Using Heron's Formula, where

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

$$s=fraca+b+c2=r+5$$

We get

$$[triangleC_1C_2C_3]=sqrt(5+r)cdot 2 cdot 3 cdot rtag2$$

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

$$6(r^2+5r)=frac254 cdot (5-r)^2 Longleftrightarrow r^2-370r+625=0$$

Solving, we get $r=5(37pm8sqrt21)$

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

Thus, $colorredr=5(37-8sqrt21)$

---

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 $$k_i=pm dfrac1r_i$$ where

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


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

So, $$k_1=-frac15, k_2=frac13 text and k_3=frac12$$

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 $r_4=dfrac1m$

share|cite|improve this answer

edited Jul 26 '16 at 14:54

answered Jul 26 '16 at 11:53

Dragonemperor42Dragonemperor42

4,0431925

$endgroup$

share|cite|improve this answer

edited Jul 26 '16 at 14:54

answered Jul 26 '16 at 11:53

Dragonemperor42Dragonemperor42

4,0431925

answered Jul 26 '16 at 11:53

Dragonemperor42Dragonemperor42

4,0431925

answered Jul 26 '16 at 11:53

Dragonemperor42Dragonemperor42

4,0431925