If $S, S_1, S_2$ be the circles of radii 5,3 and 2 respectively. If $S_1$ and $S_2$ touch externally and they...
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If $S, S_1, S_2$ be the circles of radii 5,3 and 2 respectively. If $S_1$ and $S_2$ touch externally and they touch internally with $S$. The radius of circle $S_3$ which touches externally with $S_1$ and $S_2$ and internally with $S$ is?
I tried making a diagram and figuring out, but cannot bring a relation.
geometry circle
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closed as off-topic by Saad, Paul Frost, Eevee Trainer, KReiser, Lord_Farin Dec 27 '18 at 9:05
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$begingroup$
If $S, S_1, S_2$ be the circles of radii 5,3 and 2 respectively. If $S_1$ and $S_2$ touch externally and they touch internally with $S$. The radius of circle $S_3$ which touches externally with $S_1$ and $S_2$ and internally with $S$ is?
I tried making a diagram and figuring out, but cannot bring a relation.
geometry circle
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closed as off-topic by Saad, Paul Frost, Eevee Trainer, KReiser, Lord_Farin Dec 27 '18 at 9:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Paul Frost, Eevee Trainer, KReiser, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
If $S, S_1, S_2$ be the circles of radii 5,3 and 2 respectively. If $S_1$ and $S_2$ touch externally and they touch internally with $S$. The radius of circle $S_3$ which touches externally with $S_1$ and $S_2$ and internally with $S$ is?
I tried making a diagram and figuring out, but cannot bring a relation.
geometry circle
$endgroup$
If $S, S_1, S_2$ be the circles of radii 5,3 and 2 respectively. If $S_1$ and $S_2$ touch externally and they touch internally with $S$. The radius of circle $S_3$ which touches externally with $S_1$ and $S_2$ and internally with $S$ is?
I tried making a diagram and figuring out, but cannot bring a relation.
geometry circle
geometry circle
asked Dec 26 '18 at 3:37
J. DoeJ. Doe
61
61
closed as off-topic by Saad, Paul Frost, Eevee Trainer, KReiser, Lord_Farin Dec 27 '18 at 9:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Paul Frost, Eevee Trainer, KReiser, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Paul Frost, Eevee Trainer, KReiser, Lord_Farin Dec 27 '18 at 9:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Paul Frost, Eevee Trainer, KReiser, Lord_Farin
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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Let $C_1, C_2$ be the centers of the circles $S_1, S_2$ and $C$ the center of the circle $S$. If the circle with center $O$ and radius $r$ touches the circles $S_1, S_2$ externally and $S$ internally, then we have $C_1C_2 = 5$, $OC_1 = r+2$, $OC_2=r+3$, $OC=5-r$. If in the triangle $ABC$, the point $D$ is on $BC$ such that $BD:DC = m:n$, then it is not difficult to show that
$$ (m+n)^2 AD^2 = (m+n)(m AC^2 + n AB^2) - mn BC^2$$
(See, Loney, Plane Trigonometry, Page 187, Ex 29).
Applying this to the triangle $CC_1C_2$, we have
$$ 25(5-r)^2 = 5(2(r+2)^2 + 3(r+3)^2) - 150$$
from which we get $r = 30/19$.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $C_1, C_2$ be the centers of the circles $S_1, S_2$ and $C$ the center of the circle $S$. If the circle with center $O$ and radius $r$ touches the circles $S_1, S_2$ externally and $S$ internally, then we have $C_1C_2 = 5$, $OC_1 = r+2$, $OC_2=r+3$, $OC=5-r$. If in the triangle $ABC$, the point $D$ is on $BC$ such that $BD:DC = m:n$, then it is not difficult to show that
$$ (m+n)^2 AD^2 = (m+n)(m AC^2 + n AB^2) - mn BC^2$$
(See, Loney, Plane Trigonometry, Page 187, Ex 29).
Applying this to the triangle $CC_1C_2$, we have
$$ 25(5-r)^2 = 5(2(r+2)^2 + 3(r+3)^2) - 150$$
from which we get $r = 30/19$.
$endgroup$
add a comment |
$begingroup$
Let $C_1, C_2$ be the centers of the circles $S_1, S_2$ and $C$ the center of the circle $S$. If the circle with center $O$ and radius $r$ touches the circles $S_1, S_2$ externally and $S$ internally, then we have $C_1C_2 = 5$, $OC_1 = r+2$, $OC_2=r+3$, $OC=5-r$. If in the triangle $ABC$, the point $D$ is on $BC$ such that $BD:DC = m:n$, then it is not difficult to show that
$$ (m+n)^2 AD^2 = (m+n)(m AC^2 + n AB^2) - mn BC^2$$
(See, Loney, Plane Trigonometry, Page 187, Ex 29).
Applying this to the triangle $CC_1C_2$, we have
$$ 25(5-r)^2 = 5(2(r+2)^2 + 3(r+3)^2) - 150$$
from which we get $r = 30/19$.
$endgroup$
add a comment |
$begingroup$
Let $C_1, C_2$ be the centers of the circles $S_1, S_2$ and $C$ the center of the circle $S$. If the circle with center $O$ and radius $r$ touches the circles $S_1, S_2$ externally and $S$ internally, then we have $C_1C_2 = 5$, $OC_1 = r+2$, $OC_2=r+3$, $OC=5-r$. If in the triangle $ABC$, the point $D$ is on $BC$ such that $BD:DC = m:n$, then it is not difficult to show that
$$ (m+n)^2 AD^2 = (m+n)(m AC^2 + n AB^2) - mn BC^2$$
(See, Loney, Plane Trigonometry, Page 187, Ex 29).
Applying this to the triangle $CC_1C_2$, we have
$$ 25(5-r)^2 = 5(2(r+2)^2 + 3(r+3)^2) - 150$$
from which we get $r = 30/19$.
$endgroup$
Let $C_1, C_2$ be the centers of the circles $S_1, S_2$ and $C$ the center of the circle $S$. If the circle with center $O$ and radius $r$ touches the circles $S_1, S_2$ externally and $S$ internally, then we have $C_1C_2 = 5$, $OC_1 = r+2$, $OC_2=r+3$, $OC=5-r$. If in the triangle $ABC$, the point $D$ is on $BC$ such that $BD:DC = m:n$, then it is not difficult to show that
$$ (m+n)^2 AD^2 = (m+n)(m AC^2 + n AB^2) - mn BC^2$$
(See, Loney, Plane Trigonometry, Page 187, Ex 29).
Applying this to the triangle $CC_1C_2$, we have
$$ 25(5-r)^2 = 5(2(r+2)^2 + 3(r+3)^2) - 150$$
from which we get $r = 30/19$.
answered Dec 26 '18 at 5:18
MuralidharanMuralidharan
43526
43526
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