Circle in which side of square is given [on hold]Circle Packing AlgorithmA question on circleCircle formula...
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Circle in which side of square is given [on hold]
Circle Packing AlgorithmA question on circleCircle formula given two points and a manipulable radiusarrange div elements in circle and squareCircle - finding the equationProof about the coordinates of the centre of a circle which touches another circle and the $y$-axisFinding the points of intersection on a circleFind the coordinates of the centre of a circle which is tangent to a given circleSide of a squareGiven the equation of a circle and a point on a tangent to the circle, determine the point of contact of the tangent to the circle
$begingroup$
I am not Able to proceed with this question .
circle
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
$endgroup$
put on hold as off-topic by John Omielan, mrtaurho, Thomas Shelby, Wrzlprmft, Cesareo 2 days ago
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." – John Omielan, mrtaurho, Thomas Shelby, Wrzlprmft, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I am not Able to proceed with this question .
circle
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
$endgroup$
put on hold as off-topic by John Omielan, mrtaurho, Thomas Shelby, Wrzlprmft, Cesareo 2 days ago
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." – John Omielan, mrtaurho, Thomas Shelby, Wrzlprmft, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
A great way to approach these kinds of questions is to list what you know. What information does the diagram give you? Next time, please include this information in your question itself
$endgroup$
– Aniruddh Venkatesan
2 days ago
add a comment |
$begingroup$
I am not Able to proceed with this question .
circle
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
$endgroup$
I am not Able to proceed with this question .
circle
circle
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
1354
put on hold as off-topic by John Omielan, mrtaurho, Thomas Shelby, Wrzlprmft, Cesareo 2 days ago
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." – John Omielan, mrtaurho, Thomas Shelby, Wrzlprmft, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by John Omielan, mrtaurho, Thomas Shelby, Wrzlprmft, Cesareo 2 days ago
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." – John Omielan, mrtaurho, Thomas Shelby, Wrzlprmft, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
A great way to approach these kinds of questions is to list what you know. What information does the diagram give you? Next time, please include this information in your question itself
$endgroup$
– Aniruddh Venkatesan
2 days ago
add a comment |
1
$begingroup$
A great way to approach these kinds of questions is to list what you know. What information does the diagram give you? Next time, please include this information in your question itself
$endgroup$
– Aniruddh Venkatesan
2 days ago
1
1
$begingroup$
A great way to approach these kinds of questions is to list what you know. What information does the diagram give you? Next time, please include this information in your question itself
$endgroup$
– Aniruddh Venkatesan
2 days ago
$begingroup$
A great way to approach these kinds of questions is to list what you know. What information does the diagram give you? Next time, please include this information in your question itself
$endgroup$
– Aniruddh Venkatesan
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
how can PQ be independent of a?
I am getting .6 a
answered 2 days ago
mavericmaveric
84912
$endgroup$
1
$begingroup$
I get the same thing.
$endgroup$
– marty cohen
2 days ago
add a comment |
$begingroup$
I get that
the length of "?"
is 3a/5.
This is probably the
longest possible way,
so here goes.
Let the two points
on the left and right ends
of the line labled "?"
be P and Q.
Put the origin at the base of the circle
near "a".
Let
$angle DAP = angle ADQ
=g$,
and let
$sin g = s,
cos g = c,
tan g = t
$.
The equation of
$AP$ is
$y = t(a/2-x)
$
and
the equation of
$BQ$ is
$y = t(x+a/2)
$.
These intersect when
$x = 0$
at $(0, at/2)$.
This is
the center of the circle.
Call this $R$.
The distance $AP = a$,
and
$begin{array}\
AR^2
&=(a/2)^2+(ta/2)^2\
&=(a/2)^2(1+t^2)\
&=(a/2)^2(1+(s/c)^2)\
&=(a/2)^2(1/c)^2\
text{so}\
AR
&=a/(2c)
end{array}
$
Therefore the radius
of the circle is
$a-a/(2c)
=a(1-1/(2c))
$.
But this has to be
$at/2$,
so
$at/2
=a(1-1/(2c))
$
or
$t
=2-1/c
$
or
$s/c = 2-1/c$
or
$s = 2c-1$.
Squaring,
$4c^2-4c+1
=s^2
=1-c^2
$,
so
$5c^2 = 4c$
so
$c = 0$
or
$c = 4/5$.
From this,
$s = sqrt{1-16/25}
=3/5$,
and,
indeed,
$s = 2c-1$.
Also,
$t = s/c = 3/4$.
If point $P$
is at
$(-u, v)$,
then
$(u+a/2)^2+v^2 = a^2
$
and
$v/(u+a/2)
=t$
or
$v = t(u+a/2)
$
so
$begin{array}\
a^2
&=(u+a/2)^2+t(u+a/2)^2 \
&=(u+a/2)^2(1+t^2)\
&=(u+a/2)^2(1/c^2)\
text{so}\
ac
&=u+a/2\
text{or}\
u
&=a(c-1/2)\
&=3a/10\
end{array}
$
Finally,
the length of $?$ is
$2u = 6a/10
=3a/5
$.
answered 2 days ago
marty cohenmarty cohen
74.2k549128
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
how can PQ be independent of a?
I am getting .6 a
answered 2 days ago
mavericmaveric
84912
$endgroup$
1
$begingroup$
I get the same thing.
$endgroup$
– marty cohen
2 days ago
add a comment |
$begingroup$
how can PQ be independent of a?
I am getting .6 a
answered 2 days ago
mavericmaveric
84912
$endgroup$
1
$begingroup$
I get the same thing.
$endgroup$
– marty cohen
2 days ago
add a comment |
$begingroup$
how can PQ be independent of a?
I am getting .6 a
answered 2 days ago
mavericmaveric
84912
$endgroup$
how can PQ be independent of a?
I am getting .6 a
answered 2 days ago
mavericmaveric
84912
edited 2 days ago
answered 2 days ago
mavericmaveric
84912
answered 2 days ago
mavericmaveric
84912
answered 2 days ago
mavericmaveric
84912
84912
1
$begingroup$
I get the same thing.
$endgroup$
– marty cohen
2 days ago
add a comment |
1
$begingroup$
I get the same thing.
$endgroup$
– marty cohen
2 days ago
1
1
$begingroup$
I get the same thing.
$endgroup$
– marty cohen
2 days ago
$begingroup$
I get the same thing.
$endgroup$
– marty cohen
2 days ago
add a comment |
$begingroup$
I get that
the length of "?"
is 3a/5.
This is probably the
longest possible way,
so here goes.
Let the two points
on the left and right ends
of the line labled "?"
be P and Q.
Put the origin at the base of the circle
near "a".
Let
$angle DAP = angle ADQ
=g$,
and let
$sin g = s,
cos g = c,
tan g = t
$.
The equation of
$AP$ is
$y = t(a/2-x)
$
and
the equation of
$BQ$ is
$y = t(x+a/2)
$.
These intersect when
$x = 0$
at $(0, at/2)$.
This is
the center of the circle.
Call this $R$.
The distance $AP = a$,
and
$begin{array}\
AR^2
&=(a/2)^2+(ta/2)^2\
&=(a/2)^2(1+t^2)\
&=(a/2)^2(1+(s/c)^2)\
&=(a/2)^2(1/c)^2\
text{so}\
AR
&=a/(2c)
end{array}
$
Therefore the radius
of the circle is
$a-a/(2c)
=a(1-1/(2c))
$.
But this has to be
$at/2$,
so
$at/2
=a(1-1/(2c))
$
or
$t
=2-1/c
$
or
$s/c = 2-1/c$
or
$s = 2c-1$.
Squaring,
$4c^2-4c+1
=s^2
=1-c^2
$,
so
$5c^2 = 4c$
so
$c = 0$
or
$c = 4/5$.
From this,
$s = sqrt{1-16/25}
=3/5$,
and,
indeed,
$s = 2c-1$.
Also,
$t = s/c = 3/4$.
If point $P$
is at
$(-u, v)$,
then
$(u+a/2)^2+v^2 = a^2
$
and
$v/(u+a/2)
=t$
or
$v = t(u+a/2)
$
so
$begin{array}\
a^2
&=(u+a/2)^2+t(u+a/2)^2 \
&=(u+a/2)^2(1+t^2)\
&=(u+a/2)^2(1/c^2)\
text{so}\
ac
&=u+a/2\
text{or}\
u
&=a(c-1/2)\
&=3a/10\
end{array}
$
Finally,
the length of $?$ is
$2u = 6a/10
=3a/5
$.
answered 2 days ago
marty cohenmarty cohen
74.2k549128
$endgroup$
add a comment |
$begingroup$
I get that
the length of "?"
is 3a/5.
This is probably the
longest possible way,
so here goes.
Let the two points
on the left and right ends
of the line labled "?"
be P and Q.
Put the origin at the base of the circle
near "a".
Let
$angle DAP = angle ADQ
=g$,
and let
$sin g = s,
cos g = c,
tan g = t
$.
The equation of
$AP$ is
$y = t(a/2-x)
$
and
the equation of
$BQ$ is
$y = t(x+a/2)
$.
These intersect when
$x = 0$
at $(0, at/2)$.
This is
the center of the circle.
Call this $R$.
The distance $AP = a$,
and
$begin{array}\
AR^2
&=(a/2)^2+(ta/2)^2\
&=(a/2)^2(1+t^2)\
&=(a/2)^2(1+(s/c)^2)\
&=(a/2)^2(1/c)^2\
text{so}\
AR
&=a/(2c)
end{array}
$
Therefore the radius
of the circle is
$a-a/(2c)
=a(1-1/(2c))
$.
But this has to be
$at/2$,
so
$at/2
=a(1-1/(2c))
$
or
$t
=2-1/c
$
or
$s/c = 2-1/c$
or
$s = 2c-1$.
Squaring,
$4c^2-4c+1
=s^2
=1-c^2
$,
so
$5c^2 = 4c$
so
$c = 0$
or
$c = 4/5$.
From this,
$s = sqrt{1-16/25}
=3/5$,
and,
indeed,
$s = 2c-1$.
Also,
$t = s/c = 3/4$.
If point $P$
is at
$(-u, v)$,
then
$(u+a/2)^2+v^2 = a^2
$
and
$v/(u+a/2)
=t$
or
$v = t(u+a/2)
$
so
$begin{array}\
a^2
&=(u+a/2)^2+t(u+a/2)^2 \
&=(u+a/2)^2(1+t^2)\
&=(u+a/2)^2(1/c^2)\
text{so}\
ac
&=u+a/2\
text{or}\
u
&=a(c-1/2)\
&=3a/10\
end{array}
$
Finally,
the length of $?$ is
$2u = 6a/10
=3a/5
$.
answered 2 days ago
marty cohenmarty cohen
74.2k549128
$endgroup$
add a comment |
$begingroup$
I get that
the length of "?"
is 3a/5.
This is probably the
longest possible way,
so here goes.
Let the two points
on the left and right ends
of the line labled "?"
be P and Q.
Put the origin at the base of the circle
near "a".
Let
$angle DAP = angle ADQ
=g$,
and let
$sin g = s,
cos g = c,
tan g = t
$.
The equation of
$AP$ is
$y = t(a/2-x)
$
and
the equation of
$BQ$ is
$y = t(x+a/2)
$.
These intersect when
$x = 0$
at $(0, at/2)$.
This is
the center of the circle.
Call this $R$.
The distance $AP = a$,
and
$begin{array}\
AR^2
&=(a/2)^2+(ta/2)^2\
&=(a/2)^2(1+t^2)\
&=(a/2)^2(1+(s/c)^2)\
&=(a/2)^2(1/c)^2\
text{so}\
AR
&=a/(2c)
end{array}
$
Therefore the radius
of the circle is
$a-a/(2c)
=a(1-1/(2c))
$.
But this has to be
$at/2$,
so
$at/2
=a(1-1/(2c))
$
or
$t
=2-1/c
$
or
$s/c = 2-1/c$
or
$s = 2c-1$.
Squaring,
$4c^2-4c+1
=s^2
=1-c^2
$,
so
$5c^2 = 4c$
so
$c = 0$
or
$c = 4/5$.
From this,
$s = sqrt{1-16/25}
=3/5$,
and,
indeed,
$s = 2c-1$.
Also,
$t = s/c = 3/4$.
If point $P$
is at
$(-u, v)$,
then
$(u+a/2)^2+v^2 = a^2
$
and
$v/(u+a/2)
=t$
or
$v = t(u+a/2)
$
so
$begin{array}\
a^2
&=(u+a/2)^2+t(u+a/2)^2 \
&=(u+a/2)^2(1+t^2)\
&=(u+a/2)^2(1/c^2)\
text{so}\
ac
&=u+a/2\
text{or}\
u
&=a(c-1/2)\
&=3a/10\
end{array}
$
Finally,
the length of $?$ is
$2u = 6a/10
=3a/5
$.
answered 2 days ago
marty cohenmarty cohen
74.2k549128
$endgroup$
I get that
the length of "?"
is 3a/5.
This is probably the
longest possible way,
so here goes.
Let the two points
on the left and right ends
of the line labled "?"
be P and Q.
Put the origin at the base of the circle
near "a".
Let
$angle DAP = angle ADQ
=g$,
and let
$sin g = s,
cos g = c,
tan g = t
$.
The equation of
$AP$ is
$y = t(a/2-x)
$
and
the equation of
$BQ$ is
$y = t(x+a/2)
$.
These intersect when
$x = 0$
at $(0, at/2)$.
This is
the center of the circle.
Call this $R$.
The distance $AP = a$,
and
$begin{array}\
AR^2
&=(a/2)^2+(ta/2)^2\
&=(a/2)^2(1+t^2)\
&=(a/2)^2(1+(s/c)^2)\
&=(a/2)^2(1/c)^2\
text{so}\
AR
&=a/(2c)
end{array}
$
Therefore the radius
of the circle is
$a-a/(2c)
=a(1-1/(2c))
$.
But this has to be
$at/2$,
so
$at/2
=a(1-1/(2c))
$
or
$t
=2-1/c
$
or
$s/c = 2-1/c$
or
$s = 2c-1$.
Squaring,
$4c^2-4c+1
=s^2
=1-c^2
$,
so
$5c^2 = 4c$
so
$c = 0$
or
$c = 4/5$.
From this,
$s = sqrt{1-16/25}
=3/5$,
and,
indeed,
$s = 2c-1$.
Also,
$t = s/c = 3/4$.
If point $P$
is at
$(-u, v)$,
then
$(u+a/2)^2+v^2 = a^2
$
and
$v/(u+a/2)
=t$
or
$v = t(u+a/2)
$
so
$begin{array}\
a^2
&=(u+a/2)^2+t(u+a/2)^2 \
&=(u+a/2)^2(1+t^2)\
&=(u+a/2)^2(1/c^2)\
text{so}\
ac
&=u+a/2\
text{or}\
u
&=a(c-1/2)\
&=3a/10\
end{array}
$
Finally,
the length of $?$ is
$2u = 6a/10
=3a/5
$.
answered 2 days ago
marty cohenmarty cohen
74.2k549128
answered 2 days ago
marty cohenmarty cohen
74.2k549128
answered 2 days ago
marty cohenmarty cohen
74.2k549128
answered 2 days ago
marty cohenmarty cohen
74.2k549128
74.2k549128
add a comment |
add a comment |
1
$begingroup$
A great way to approach these kinds of questions is to list what you know. What information does the diagram give you? Next time, please include this information in your question itself
$endgroup$
– Aniruddh Venkatesan
2 days ago