Question about inversion images
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"Let $W$ be a circle with center $O$ and radius $r$. Let $S$ be a point outside
the circle. Let $l_1$ and $l_2$ be two tangent lines to the circle $W$ passing through the point $S$ . Let $T_1$ and $T_2$ be the points of tangency of these lines with the circle. Show that
the point of intersection of the segments $OS$ and $T_1$$T_2$ is the image of $S$ under inversion in $W$"
I really do not know how to get started with this question, apart from knowing that $OS$*$OS' = R^2$ where $R$ is the radius of the new circle. Help?
geometry linear-transformations euclidean-geometry transformation geometric-transformation
edited Nov 28 at 21:07
greedoid
36.7k114593
asked Nov 28 at 21:01
helpneeded
897
add a comment |
up vote
0
down vote
favorite
"Let $W$ be a circle with center $O$ and radius $r$. Let $S$ be a point outside
the circle. Let $l_1$ and $l_2$ be two tangent lines to the circle $W$ passing through the point $S$ . Let $T_1$ and $T_2$ be the points of tangency of these lines with the circle. Show that
the point of intersection of the segments $OS$ and $T_1$$T_2$ is the image of $S$ under inversion in $W$"
I really do not know how to get started with this question, apart from knowing that $OS$*$OS' = R^2$ where $R$ is the radius of the new circle. Help?
geometry linear-transformations euclidean-geometry transformation geometric-transformation
edited Nov 28 at 21:07
greedoid
36.7k114593
asked Nov 28 at 21:01
helpneeded
897
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
"Let $W$ be a circle with center $O$ and radius $r$. Let $S$ be a point outside
the circle. Let $l_1$ and $l_2$ be two tangent lines to the circle $W$ passing through the point $S$ . Let $T_1$ and $T_2$ be the points of tangency of these lines with the circle. Show that
the point of intersection of the segments $OS$ and $T_1$$T_2$ is the image of $S$ under inversion in $W$"
I really do not know how to get started with this question, apart from knowing that $OS$*$OS' = R^2$ where $R$ is the radius of the new circle. Help?
geometry linear-transformations euclidean-geometry transformation geometric-transformation
edited Nov 28 at 21:07
greedoid
36.7k114593
asked Nov 28 at 21:01
helpneeded
897
"Let $W$ be a circle with center $O$ and radius $r$. Let $S$ be a point outside
the circle. Let $l_1$ and $l_2$ be two tangent lines to the circle $W$ passing through the point $S$ . Let $T_1$ and $T_2$ be the points of tangency of these lines with the circle. Show that
the point of intersection of the segments $OS$ and $T_1$$T_2$ is the image of $S$ under inversion in $W$"
I really do not know how to get started with this question, apart from knowing that $OS$*$OS' = R^2$ where $R$ is the radius of the new circle. Help?
geometry linear-transformations euclidean-geometry transformation geometric-transformation
geometry linear-transformations euclidean-geometry transformation geometric-transformation
edited Nov 28 at 21:07
greedoid
36.7k114593
asked Nov 28 at 21:01
helpneeded
897
edited Nov 28 at 21:07
greedoid
36.7k114593
asked Nov 28 at 21:01
helpneeded
897
edited Nov 28 at 21:07
greedoid
36.7k114593
edited Nov 28 at 21:07
greedoid
36.7k114593
edited Nov 28 at 21:07
greedoid
36.7k114593
36.7k114593
asked Nov 28 at 21:01
helpneeded
897
asked Nov 28 at 21:01
helpneeded
897
asked Nov 28 at 21:01
helpneeded
897
897
add a comment |
add a comment |
1 Answer
1
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oldest
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1
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accepted
Let $X$ be an intersection point of $T_1T_2$ and $OS$. Since circle around $XT_1S$ touches $OT_1$ we have, by the PoP of $O$ with respect to that new circle:$$ r^2 =OT_1^2 = OC cdot OS$$
and we are done.
answered Nov 28 at 21:07
greedoid
36.7k114593
What does "PoP" stand for?
– helpneeded
Nov 29 at 0:27
Pover of the point
– greedoid
Nov 29 at 9:32
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $X$ be an intersection point of $T_1T_2$ and $OS$. Since circle around $XT_1S$ touches $OT_1$ we have, by the PoP of $O$ with respect to that new circle:$$ r^2 =OT_1^2 = OC cdot OS$$
and we are done.
answered Nov 28 at 21:07
greedoid
36.7k114593
What does "PoP" stand for?
– helpneeded
Nov 29 at 0:27
Pover of the point
– greedoid
Nov 29 at 9:32
add a comment |
up vote
1
down vote
accepted
Let $X$ be an intersection point of $T_1T_2$ and $OS$. Since circle around $XT_1S$ touches $OT_1$ we have, by the PoP of $O$ with respect to that new circle:$$ r^2 =OT_1^2 = OC cdot OS$$
and we are done.
answered Nov 28 at 21:07
greedoid
36.7k114593
What does "PoP" stand for?
– helpneeded
Nov 29 at 0:27
Pover of the point
– greedoid
Nov 29 at 9:32
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $X$ be an intersection point of $T_1T_2$ and $OS$. Since circle around $XT_1S$ touches $OT_1$ we have, by the PoP of $O$ with respect to that new circle:$$ r^2 =OT_1^2 = OC cdot OS$$
and we are done.
answered Nov 28 at 21:07
greedoid
36.7k114593
Let $X$ be an intersection point of $T_1T_2$ and $OS$. Since circle around $XT_1S$ touches $OT_1$ we have, by the PoP of $O$ with respect to that new circle:$$ r^2 =OT_1^2 = OC cdot OS$$
and we are done.
answered Nov 28 at 21:07
greedoid
36.7k114593
answered Nov 28 at 21:07
greedoid
36.7k114593
answered Nov 28 at 21:07
greedoid
36.7k114593
answered Nov 28 at 21:07
greedoid
36.7k114593
36.7k114593
What does "PoP" stand for?
– helpneeded
Nov 29 at 0:27
Pover of the point
– greedoid
Nov 29 at 9:32
add a comment |
What does "PoP" stand for?
– helpneeded
Nov 29 at 0:27
Pover of the point
– greedoid
Nov 29 at 9:32
What does "PoP" stand for?
– helpneeded
Nov 29 at 0:27
What does "PoP" stand for?
– helpneeded
Nov 29 at 0:27
Pover of the point
– greedoid
Nov 29 at 9:32
Pover of the point
– greedoid
Nov 29 at 9:32
add a comment |
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