Mixing
P and Q are 4 units apart and on the same side of variable straight line which touches a circle
$begingroup$
My approach: I took a variable line ax+by+c=0 and took random points like (2,0) and (6,0). The I applied the formula for perpendicular distance and put it in the condition that PM+3QN=4.
but this way there are three variable and one equation by which I am unable to get the equation of the line and satisfy it with the tangency condition of the circle that perpendicular distance from centre to this line is equal to radius but this way more and more number of variables are introduced and lesser number of equations.
combinatorics geometry circles coordinate-systems
asked Feb 1 at 4:51
Sarah janeSarah jane
1897
$endgroup$
add a comment |
$begingroup$
My approach: I took a variable line ax+by+c=0 and took random points like (2,0) and (6,0). The I applied the formula for perpendicular distance and put it in the condition that PM+3QN=4.
but this way there are three variable and one equation by which I am unable to get the equation of the line and satisfy it with the tangency condition of the circle that perpendicular distance from centre to this line is equal to radius but this way more and more number of variables are introduced and lesser number of equations.
combinatorics geometry circles coordinate-systems
asked Feb 1 at 4:51
Sarah janeSarah jane
1897
$endgroup$
1
$begingroup$
Did you try to draw a graph? How about you put the two points in a more simmetrical position. Like $(-2,0)$ and $(2,0)$. This might help. Also. Note that $PM$ and $QN$ are parallel. What quadrilateral is $PQMN$?
$endgroup$
– Matteo
Feb 1 at 9:31
add a comment |
$begingroup$
My approach: I took a variable line ax+by+c=0 and took random points like (2,0) and (6,0). The I applied the formula for perpendicular distance and put it in the condition that PM+3QN=4.
but this way there are three variable and one equation by which I am unable to get the equation of the line and satisfy it with the tangency condition of the circle that perpendicular distance from centre to this line is equal to radius but this way more and more number of variables are introduced and lesser number of equations.
combinatorics geometry circles coordinate-systems
asked Feb 1 at 4:51
Sarah janeSarah jane
1897
$endgroup$
My approach: I took a variable line ax+by+c=0 and took random points like (2,0) and (6,0). The I applied the formula for perpendicular distance and put it in the condition that PM+3QN=4.
but this way there are three variable and one equation by which I am unable to get the equation of the line and satisfy it with the tangency condition of the circle that perpendicular distance from centre to this line is equal to radius but this way more and more number of variables are introduced and lesser number of equations.
combinatorics geometry circles coordinate-systems
combinatorics geometry circles coordinate-systems
asked Feb 1 at 4:51
Sarah janeSarah jane
1897
asked Feb 1 at 4:51
Sarah janeSarah jane
1897
asked Feb 1 at 4:51
Sarah janeSarah jane
1897
asked Feb 1 at 4:51
Sarah janeSarah jane
1897
asked Feb 1 at 4:51
Sarah janeSarah jane
1897
1897
1
$begingroup$
Did you try to draw a graph? How about you put the two points in a more simmetrical position. Like $(-2,0)$ and $(2,0)$. This might help. Also. Note that $PM$ and $QN$ are parallel. What quadrilateral is $PQMN$?
$endgroup$
– Matteo
Feb 1 at 9:31
add a comment |
1
$begingroup$
Did you try to draw a graph? How about you put the two points in a more simmetrical position. Like $(-2,0)$ and $(2,0)$. This might help. Also. Note that $PM$ and $QN$ are parallel. What quadrilateral is $PQMN$?
$endgroup$
– Matteo
Feb 1 at 9:31
1
1
$begingroup$
Did you try to draw a graph? How about you put the two points in a more simmetrical position. Like $(-2,0)$ and $(2,0)$. This might help. Also. Note that $PM$ and $QN$ are parallel. What quadrilateral is $PQMN$?
$endgroup$
– Matteo
Feb 1 at 9:31
$begingroup$
Did you try to draw a graph? How about you put the two points in a more simmetrical position. Like $(-2,0)$ and $(2,0)$. This might help. Also. Note that $PM$ and $QN$ are parallel. What quadrilateral is $PQMN$?
$endgroup$
– Matteo
Feb 1 at 9:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For an arbitrary line $L$ extend $MN$ until it meets $PQ$ at point $A$. Introduce angle $alpha=angle MAP$ and length $b=AQ$. The following is obvious:
$$PM=(4+b)sinalpha$$
$$QN=bsinalpha$$
We know that $PM+3QN=4$. It means that:
$$(4+b)sinalpha+3bsinalpha=4$$
$$sinalpha=frac1{b+1}tag{1}$$
In other words, parameters $b$ and $alpha$ are not independent. You can pick either $b$ or $alpha$ freely and the other one can be calculated from (1).
Let us now calculate distance $d$ for an arbitrary point $Rin{PQ}$ such that $QR=x$:
$$d=(b+x)sinalpha=frac{b+x}{b+1}tag{2}$$
So for an arbitrary point $R$ and arbitrary line $L$, the distance from $R$ to $L$ is a function of $x$ (which represents position of point $R$ along the line $PQ$) and $b$ (which can be picked freely).
But there is one special point on segment PQ. If you put $x=1$ into (2) you get $d=1$ for all possible values of $b$. In other words, the point $R$ such that $PR=3,QR=1$ has the same distance $(d=1)$ from an arbitrary line $L$. So if you draw a circle around point $R$ with radius 1 it will be tangent to the line $L$.
It means that (A) and (B) are correct.
answered Feb 1 at 14:28
OldboyOldboy
9,42911138
$endgroup$
$begingroup$
But in question it is given PM+3QN=4
$endgroup$
– Sarah jane
Feb 2 at 17:01
$begingroup$
@Sarahjane Thanks, I have made a typo. Corrected.
$endgroup$
– Oldboy
Feb 2 at 23:54
$begingroup$
still it's a good observation. thanks a lot
$endgroup$
– Sarah jane
Feb 5 at 7:48
add a comment |
$begingroup$
If you call $theta$ the angle $angle MPQ$, then you have
$$overline{PM} = 4-3cdotoverline{QN} = 4costheta-overline{QN},$$
which gives you
begin{equation}overline{QN} = 1-costhetatag{1}label{uno}end{equation}
and consequently
begin{equation}overline{PM} = 1+3costheta.tag{2}label{due}end{equation}
Then it is useful to put $P$ in the origin of the axes.
Note that
$$overline{PH} = 3costheta$$
and, thus, from eqref{due},
$$overline{HM} = 1.$$
Then if you draw the line $HK$ perpendicular to $PM$ you get
$$overline{KN} = 1,$$
and
$$overline{KQ} = costheta.$$
Again, as in eqref{uno},
$$overline{QN} = 1-costheta.$$
As a consequence, the distance of the line $MN$ from $G(3,0)$ will be constant, and the line will be tangent to the circle centered in $G$ with radius $1$.
answered Feb 1 at 14:34
MatteoMatteo
1,3121313
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
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active
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active
oldest
votes
$begingroup$
For an arbitrary line $L$ extend $MN$ until it meets $PQ$ at point $A$. Introduce angle $alpha=angle MAP$ and length $b=AQ$. The following is obvious:
$$PM=(4+b)sinalpha$$
$$QN=bsinalpha$$
We know that $PM+3QN=4$. It means that:
$$(4+b)sinalpha+3bsinalpha=4$$
$$sinalpha=frac1{b+1}tag{1}$$
In other words, parameters $b$ and $alpha$ are not independent. You can pick either $b$ or $alpha$ freely and the other one can be calculated from (1).
Let us now calculate distance $d$ for an arbitrary point $Rin{PQ}$ such that $QR=x$:
$$d=(b+x)sinalpha=frac{b+x}{b+1}tag{2}$$
So for an arbitrary point $R$ and arbitrary line $L$, the distance from $R$ to $L$ is a function of $x$ (which represents position of point $R$ along the line $PQ$) and $b$ (which can be picked freely).
But there is one special point on segment PQ. If you put $x=1$ into (2) you get $d=1$ for all possible values of $b$. In other words, the point $R$ such that $PR=3,QR=1$ has the same distance $(d=1)$ from an arbitrary line $L$. So if you draw a circle around point $R$ with radius 1 it will be tangent to the line $L$.
It means that (A) and (B) are correct.
answered Feb 1 at 14:28
OldboyOldboy
9,42911138
$endgroup$
$begingroup$
But in question it is given PM+3QN=4
$endgroup$
– Sarah jane
Feb 2 at 17:01
$begingroup$
@Sarahjane Thanks, I have made a typo. Corrected.
$endgroup$
– Oldboy
Feb 2 at 23:54
$begingroup$
still it's a good observation. thanks a lot
$endgroup$
– Sarah jane
Feb 5 at 7:48
add a comment |
$begingroup$
For an arbitrary line $L$ extend $MN$ until it meets $PQ$ at point $A$. Introduce angle $alpha=angle MAP$ and length $b=AQ$. The following is obvious:
$$PM=(4+b)sinalpha$$
$$QN=bsinalpha$$
We know that $PM+3QN=4$. It means that:
$$(4+b)sinalpha+3bsinalpha=4$$
$$sinalpha=frac1{b+1}tag{1}$$
In other words, parameters $b$ and $alpha$ are not independent. You can pick either $b$ or $alpha$ freely and the other one can be calculated from (1).
Let us now calculate distance $d$ for an arbitrary point $Rin{PQ}$ such that $QR=x$:
$$d=(b+x)sinalpha=frac{b+x}{b+1}tag{2}$$
So for an arbitrary point $R$ and arbitrary line $L$, the distance from $R$ to $L$ is a function of $x$ (which represents position of point $R$ along the line $PQ$) and $b$ (which can be picked freely).
But there is one special point on segment PQ. If you put $x=1$ into (2) you get $d=1$ for all possible values of $b$. In other words, the point $R$ such that $PR=3,QR=1$ has the same distance $(d=1)$ from an arbitrary line $L$. So if you draw a circle around point $R$ with radius 1 it will be tangent to the line $L$.
It means that (A) and (B) are correct.
answered Feb 1 at 14:28
OldboyOldboy
9,42911138
$endgroup$
$begingroup$
But in question it is given PM+3QN=4
$endgroup$
– Sarah jane
Feb 2 at 17:01
$begingroup$
@Sarahjane Thanks, I have made a typo. Corrected.
$endgroup$
– Oldboy
Feb 2 at 23:54
$begingroup$
still it's a good observation. thanks a lot
$endgroup$
– Sarah jane
Feb 5 at 7:48
add a comment |
$begingroup$
For an arbitrary line $L$ extend $MN$ until it meets $PQ$ at point $A$. Introduce angle $alpha=angle MAP$ and length $b=AQ$. The following is obvious:
$$PM=(4+b)sinalpha$$
$$QN=bsinalpha$$
We know that $PM+3QN=4$. It means that:
$$(4+b)sinalpha+3bsinalpha=4$$
$$sinalpha=frac1{b+1}tag{1}$$
In other words, parameters $b$ and $alpha$ are not independent. You can pick either $b$ or $alpha$ freely and the other one can be calculated from (1).
Let us now calculate distance $d$ for an arbitrary point $Rin{PQ}$ such that $QR=x$:
$$d=(b+x)sinalpha=frac{b+x}{b+1}tag{2}$$
So for an arbitrary point $R$ and arbitrary line $L$, the distance from $R$ to $L$ is a function of $x$ (which represents position of point $R$ along the line $PQ$) and $b$ (which can be picked freely).
But there is one special point on segment PQ. If you put $x=1$ into (2) you get $d=1$ for all possible values of $b$. In other words, the point $R$ such that $PR=3,QR=1$ has the same distance $(d=1)$ from an arbitrary line $L$. So if you draw a circle around point $R$ with radius 1 it will be tangent to the line $L$.
It means that (A) and (B) are correct.
answered Feb 1 at 14:28
OldboyOldboy
9,42911138
$endgroup$
For an arbitrary line $L$ extend $MN$ until it meets $PQ$ at point $A$. Introduce angle $alpha=angle MAP$ and length $b=AQ$. The following is obvious:
$$PM=(4+b)sinalpha$$
$$QN=bsinalpha$$
We know that $PM+3QN=4$. It means that:
$$(4+b)sinalpha+3bsinalpha=4$$
$$sinalpha=frac1{b+1}tag{1}$$
In other words, parameters $b$ and $alpha$ are not independent. You can pick either $b$ or $alpha$ freely and the other one can be calculated from (1).
Let us now calculate distance $d$ for an arbitrary point $Rin{PQ}$ such that $QR=x$:
$$d=(b+x)sinalpha=frac{b+x}{b+1}tag{2}$$
So for an arbitrary point $R$ and arbitrary line $L$, the distance from $R$ to $L$ is a function of $x$ (which represents position of point $R$ along the line $PQ$) and $b$ (which can be picked freely).
But there is one special point on segment PQ. If you put $x=1$ into (2) you get $d=1$ for all possible values of $b$. In other words, the point $R$ such that $PR=3,QR=1$ has the same distance $(d=1)$ from an arbitrary line $L$. So if you draw a circle around point $R$ with radius 1 it will be tangent to the line $L$.
It means that (A) and (B) are correct.
answered Feb 1 at 14:28
OldboyOldboy
9,42911138
edited Feb 2 at 23:53
answered Feb 1 at 14:28
OldboyOldboy
9,42911138
answered Feb 1 at 14:28
OldboyOldboy
9,42911138
answered Feb 1 at 14:28
OldboyOldboy
9,42911138
9,42911138
$begingroup$
But in question it is given PM+3QN=4
$endgroup$
– Sarah jane
Feb 2 at 17:01
$begingroup$
@Sarahjane Thanks, I have made a typo. Corrected.
$endgroup$
– Oldboy
Feb 2 at 23:54
$begingroup$
still it's a good observation. thanks a lot
$endgroup$
– Sarah jane
Feb 5 at 7:48
add a comment |
$begingroup$
But in question it is given PM+3QN=4
$endgroup$
– Sarah jane
Feb 2 at 17:01
$begingroup$
@Sarahjane Thanks, I have made a typo. Corrected.
$endgroup$
– Oldboy
Feb 2 at 23:54
$begingroup$
still it's a good observation. thanks a lot
$endgroup$
– Sarah jane
Feb 5 at 7:48
$begingroup$
But in question it is given PM+3QN=4
$endgroup$
– Sarah jane
Feb 2 at 17:01
$begingroup$
But in question it is given PM+3QN=4
$endgroup$
– Sarah jane
Feb 2 at 17:01
$begingroup$
@Sarahjane Thanks, I have made a typo. Corrected.
$endgroup$
– Oldboy
Feb 2 at 23:54
$begingroup$
@Sarahjane Thanks, I have made a typo. Corrected.
$endgroup$
– Oldboy
Feb 2 at 23:54
$begingroup$
still it's a good observation. thanks a lot
$endgroup$
– Sarah jane
Feb 5 at 7:48
$begingroup$
still it's a good observation. thanks a lot
$endgroup$
– Sarah jane
Feb 5 at 7:48
add a comment |
$begingroup$
If you call $theta$ the angle $angle MPQ$, then you have
$$overline{PM} = 4-3cdotoverline{QN} = 4costheta-overline{QN},$$
which gives you
begin{equation}overline{QN} = 1-costhetatag{1}label{uno}end{equation}
and consequently
begin{equation}overline{PM} = 1+3costheta.tag{2}label{due}end{equation}
Then it is useful to put $P$ in the origin of the axes.
Note that
$$overline{PH} = 3costheta$$
and, thus, from eqref{due},
$$overline{HM} = 1.$$
Then if you draw the line $HK$ perpendicular to $PM$ you get
$$overline{KN} = 1,$$
and
$$overline{KQ} = costheta.$$
Again, as in eqref{uno},
$$overline{QN} = 1-costheta.$$
As a consequence, the distance of the line $MN$ from $G(3,0)$ will be constant, and the line will be tangent to the circle centered in $G$ with radius $1$.
answered Feb 1 at 14:34
MatteoMatteo
1,3121313
$endgroup$
add a comment |
$begingroup$
If you call $theta$ the angle $angle MPQ$, then you have
$$overline{PM} = 4-3cdotoverline{QN} = 4costheta-overline{QN},$$
which gives you
begin{equation}overline{QN} = 1-costhetatag{1}label{uno}end{equation}
and consequently
begin{equation}overline{PM} = 1+3costheta.tag{2}label{due}end{equation}
Then it is useful to put $P$ in the origin of the axes.
Note that
$$overline{PH} = 3costheta$$
and, thus, from eqref{due},
$$overline{HM} = 1.$$
Then if you draw the line $HK$ perpendicular to $PM$ you get
$$overline{KN} = 1,$$
and
$$overline{KQ} = costheta.$$
Again, as in eqref{uno},
$$overline{QN} = 1-costheta.$$
As a consequence, the distance of the line $MN$ from $G(3,0)$ will be constant, and the line will be tangent to the circle centered in $G$ with radius $1$.
answered Feb 1 at 14:34
MatteoMatteo
1,3121313
$endgroup$
add a comment |
$begingroup$
If you call $theta$ the angle $angle MPQ$, then you have
$$overline{PM} = 4-3cdotoverline{QN} = 4costheta-overline{QN},$$
which gives you
begin{equation}overline{QN} = 1-costhetatag{1}label{uno}end{equation}
and consequently
begin{equation}overline{PM} = 1+3costheta.tag{2}label{due}end{equation}
Then it is useful to put $P$ in the origin of the axes.
Note that
$$overline{PH} = 3costheta$$
and, thus, from eqref{due},
$$overline{HM} = 1.$$
Then if you draw the line $HK$ perpendicular to $PM$ you get
$$overline{KN} = 1,$$
and
$$overline{KQ} = costheta.$$
Again, as in eqref{uno},
$$overline{QN} = 1-costheta.$$
As a consequence, the distance of the line $MN$ from $G(3,0)$ will be constant, and the line will be tangent to the circle centered in $G$ with radius $1$.
answered Feb 1 at 14:34
MatteoMatteo
1,3121313
$endgroup$
If you call $theta$ the angle $angle MPQ$, then you have
$$overline{PM} = 4-3cdotoverline{QN} = 4costheta-overline{QN},$$
which gives you
begin{equation}overline{QN} = 1-costhetatag{1}label{uno}end{equation}
and consequently
begin{equation}overline{PM} = 1+3costheta.tag{2}label{due}end{equation}
Then it is useful to put $P$ in the origin of the axes.
Note that
$$overline{PH} = 3costheta$$
and, thus, from eqref{due},
$$overline{HM} = 1.$$
Then if you draw the line $HK$ perpendicular to $PM$ you get
$$overline{KN} = 1,$$
and
$$overline{KQ} = costheta.$$
Again, as in eqref{uno},
$$overline{QN} = 1-costheta.$$
As a consequence, the distance of the line $MN$ from $G(3,0)$ will be constant, and the line will be tangent to the circle centered in $G$ with radius $1$.
answered Feb 1 at 14:34
MatteoMatteo
1,3121313
edited Feb 1 at 19:42
answered Feb 1 at 14:34
MatteoMatteo
1,3121313
answered Feb 1 at 14:34
MatteoMatteo
1,3121313
answered Feb 1 at 14:34
MatteoMatteo
1,3121313
1,3121313
add a comment |
add a comment |
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$begingroup$
Did you try to draw a graph? How about you put the two points in a more simmetrical position. Like $(-2,0)$ and $(2,0)$. This might help. Also. Note that $PM$ and $QN$ are parallel. What quadrilateral is $PQMN$?
$endgroup$
– Matteo
Feb 1 at 9:31