P is a point inside a circle and A is a point on the circumference.Find the radius of the circle.
$begingroup$
P is a point inside a circle and A is a point on the circumference. The minimum distance between A and P is 2 cm and the maximum distance between A and P is 8 cm. Find the radius of the circle.
I think the radius must be greater than 4 cm. But next how to proceed ??
circle
asked Jan 24 at 10:34
Subham SenSubham Sen
31
$endgroup$
add a comment |
$begingroup$
P is a point inside a circle and A is a point on the circumference. The minimum distance between A and P is 2 cm and the maximum distance between A and P is 8 cm. Find the radius of the circle.
I think the radius must be greater than 4 cm. But next how to proceed ??
circle
asked Jan 24 at 10:34
Subham SenSubham Sen
31
$endgroup$
$begingroup$
Is there a rule as to how $P$ is chosen other than it being inside the circle? If not, shouldn't the minimum distance between $A$ and $P$ be $0$? What makes you think the radius is $>4$?
$endgroup$
– Shubham Johri
Jan 24 at 11:09
$begingroup$
Are you familiar with triangular inequality?
$endgroup$
– Matteo
Jan 24 at 11:16
$begingroup$
@ShubhamJohri Minimum AP=2 cm is given in the question. Since P is inside a circle and maximum AP=8 cm, so the diameter is greater than 8 cm and therefore, radius > 4 cm.
$endgroup$
– Subham Sen
Jan 24 at 12:33
add a comment |
$begingroup$
P is a point inside a circle and A is a point on the circumference. The minimum distance between A and P is 2 cm and the maximum distance between A and P is 8 cm. Find the radius of the circle.
I think the radius must be greater than 4 cm. But next how to proceed ??
circle
asked Jan 24 at 10:34
Subham SenSubham Sen
31
$endgroup$
P is a point inside a circle and A is a point on the circumference. The minimum distance between A and P is 2 cm and the maximum distance between A and P is 8 cm. Find the radius of the circle.
I think the radius must be greater than 4 cm. But next how to proceed ??
circle
circle
asked Jan 24 at 10:34
Subham SenSubham Sen
31
asked Jan 24 at 10:34
Subham SenSubham Sen
31
asked Jan 24 at 10:34
Subham SenSubham Sen
31
asked Jan 24 at 10:34
Subham SenSubham Sen
31
asked Jan 24 at 10:34
Subham SenSubham Sen
31
31
$begingroup$
Is there a rule as to how $P$ is chosen other than it being inside the circle? If not, shouldn't the minimum distance between $A$ and $P$ be $0$? What makes you think the radius is $>4$?
$endgroup$
– Shubham Johri
Jan 24 at 11:09
$begingroup$
Are you familiar with triangular inequality?
$endgroup$
– Matteo
Jan 24 at 11:16
$begingroup$
@ShubhamJohri Minimum AP=2 cm is given in the question. Since P is inside a circle and maximum AP=8 cm, so the diameter is greater than 8 cm and therefore, radius > 4 cm.
$endgroup$
– Subham Sen
Jan 24 at 12:33
add a comment |
$begingroup$
Is there a rule as to how $P$ is chosen other than it being inside the circle? If not, shouldn't the minimum distance between $A$ and $P$ be $0$? What makes you think the radius is $>4$?
$endgroup$
– Shubham Johri
Jan 24 at 11:09
$begingroup$
Are you familiar with triangular inequality?
$endgroup$
– Matteo
Jan 24 at 11:16
$begingroup$
@ShubhamJohri Minimum AP=2 cm is given in the question. Since P is inside a circle and maximum AP=8 cm, so the diameter is greater than 8 cm and therefore, radius > 4 cm.
$endgroup$
– Subham Sen
Jan 24 at 12:33
$begingroup$
Is there a rule as to how $P$ is chosen other than it being inside the circle? If not, shouldn't the minimum distance between $A$ and $P$ be $0$? What makes you think the radius is $>4$?
$endgroup$
– Shubham Johri
Jan 24 at 11:09
$begingroup$
Is there a rule as to how $P$ is chosen other than it being inside the circle? If not, shouldn't the minimum distance between $A$ and $P$ be $0$? What makes you think the radius is $>4$?
$endgroup$
– Shubham Johri
Jan 24 at 11:09
$begingroup$
Are you familiar with triangular inequality?
$endgroup$
– Matteo
Jan 24 at 11:16
$begingroup$
Are you familiar with triangular inequality?
$endgroup$
– Matteo
Jan 24 at 11:16
$begingroup$
@ShubhamJohri Minimum AP=2 cm is given in the question. Since P is inside a circle and maximum AP=8 cm, so the diameter is greater than 8 cm and therefore, radius > 4 cm.
$endgroup$
– Subham Sen
Jan 24 at 12:33
$begingroup$
@ShubhamJohri Minimum AP=2 cm is given in the question. Since P is inside a circle and maximum AP=8 cm, so the diameter is greater than 8 cm and therefore, radius > 4 cm.
$endgroup$
– Subham Sen
Jan 24 at 12:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can use triangular inequality to show that the minimum and maximum distances are attained when the point $A$ is on the diameter passing through $P$ (see Figure in the bottom).
For example,
$$A'P+OP = r,$$
where $r$ is the circle radius, and, by triangular inequality on triangle $AOP$,
$$AP+OP geq r,$$
so that
$$AP geq A'P.$$
So no matter where you take $A$ the distance from $P$ will be greater than that of $A'$ from $P$. Thus
$$A'P=2 mbox{cm}.$$
Similarly, you can express $A''P$ as
$$A''P = r + OP,$$
whereas, again for triangular inequality on $AOP$,
$$AP leq r + OP,$$
which yields
$$ AP leq A''P.$$
In conclusion $A''P = 8$ cm and the radius is $r = frac{A'P + A''P}{2}=5$ cm.
$hskip1.5in$
answered Jan 24 at 11:43
MatteoMatteo
888312
$endgroup$
$begingroup$
Ok. then while doing same with A'',, will i take triangle AOA'' ?? Is AP=2 cm ??
$endgroup$
– Subham Sen
Jan 24 at 12:22
$begingroup$
@SubhamSen: $AP$ is greater than $A'P$, as we just demonstrated, right? Since $2$ cm is the minimum distance, then which segment is $2$ cm long? For the other question: try expressing $A''P$ in terms of the radius.
$endgroup$
– Matteo
Jan 24 at 13:04
$begingroup$
@SubhamSen added the conclusion in the text.
$endgroup$
– Matteo
Jan 24 at 14:21
$begingroup$
a> I couldn't get how A'P=2 cm and A''P=8 cm ??
$endgroup$
– Subham Sen
Jan 24 at 18:53
$begingroup$
b> Why are you saying, " no matter where you take A the distance from P will be greater than that of A' " ??
$endgroup$
– Subham Sen
Jan 24 at 18:56
|
show 2 more comments
$begingroup$
The minimum distance from the point on the circumference will be at a right angle to the edge of the circle and the maximum distance will pass through the center point and also be at a right angle. Adding to two distances will make the diameter of the circle: $8+2 = 10$cm, therefore the radius is $5$ cm.
edited Jan 24 at 14:08
Namaste
1
answered Jan 24 at 13:56
JamesJames
1
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use triangular inequality to show that the minimum and maximum distances are attained when the point $A$ is on the diameter passing through $P$ (see Figure in the bottom).
For example,
$$A'P+OP = r,$$
where $r$ is the circle radius, and, by triangular inequality on triangle $AOP$,
$$AP+OP geq r,$$
so that
$$AP geq A'P.$$
So no matter where you take $A$ the distance from $P$ will be greater than that of $A'$ from $P$. Thus
$$A'P=2 mbox{cm}.$$
Similarly, you can express $A''P$ as
$$A''P = r + OP,$$
whereas, again for triangular inequality on $AOP$,
$$AP leq r + OP,$$
which yields
$$ AP leq A''P.$$
In conclusion $A''P = 8$ cm and the radius is $r = frac{A'P + A''P}{2}=5$ cm.
$hskip1.5in$
answered Jan 24 at 11:43
MatteoMatteo
888312
$endgroup$
$begingroup$
Ok. then while doing same with A'',, will i take triangle AOA'' ?? Is AP=2 cm ??
$endgroup$
– Subham Sen
Jan 24 at 12:22
$begingroup$
@SubhamSen: $AP$ is greater than $A'P$, as we just demonstrated, right? Since $2$ cm is the minimum distance, then which segment is $2$ cm long? For the other question: try expressing $A''P$ in terms of the radius.
$endgroup$
– Matteo
Jan 24 at 13:04
$begingroup$
@SubhamSen added the conclusion in the text.
$endgroup$
– Matteo
Jan 24 at 14:21
$begingroup$
a> I couldn't get how A'P=2 cm and A''P=8 cm ??
$endgroup$
– Subham Sen
Jan 24 at 18:53
$begingroup$
b> Why are you saying, " no matter where you take A the distance from P will be greater than that of A' " ??
$endgroup$
– Subham Sen
Jan 24 at 18:56
|
show 2 more comments
$begingroup$
You can use triangular inequality to show that the minimum and maximum distances are attained when the point $A$ is on the diameter passing through $P$ (see Figure in the bottom).
For example,
$$A'P+OP = r,$$
where $r$ is the circle radius, and, by triangular inequality on triangle $AOP$,
$$AP+OP geq r,$$
so that
$$AP geq A'P.$$
So no matter where you take $A$ the distance from $P$ will be greater than that of $A'$ from $P$. Thus
$$A'P=2 mbox{cm}.$$
Similarly, you can express $A''P$ as
$$A''P = r + OP,$$
whereas, again for triangular inequality on $AOP$,
$$AP leq r + OP,$$
which yields
$$ AP leq A''P.$$
In conclusion $A''P = 8$ cm and the radius is $r = frac{A'P + A''P}{2}=5$ cm.
$hskip1.5in$
answered Jan 24 at 11:43
MatteoMatteo
888312
$endgroup$
$begingroup$
Ok. then while doing same with A'',, will i take triangle AOA'' ?? Is AP=2 cm ??
$endgroup$
– Subham Sen
Jan 24 at 12:22
$begingroup$
@SubhamSen: $AP$ is greater than $A'P$, as we just demonstrated, right? Since $2$ cm is the minimum distance, then which segment is $2$ cm long? For the other question: try expressing $A''P$ in terms of the radius.
$endgroup$
– Matteo
Jan 24 at 13:04
$begingroup$
@SubhamSen added the conclusion in the text.
$endgroup$
– Matteo
Jan 24 at 14:21
$begingroup$
a> I couldn't get how A'P=2 cm and A''P=8 cm ??
$endgroup$
– Subham Sen
Jan 24 at 18:53
$begingroup$
b> Why are you saying, " no matter where you take A the distance from P will be greater than that of A' " ??
$endgroup$
– Subham Sen
Jan 24 at 18:56
|
show 2 more comments
$begingroup$
You can use triangular inequality to show that the minimum and maximum distances are attained when the point $A$ is on the diameter passing through $P$ (see Figure in the bottom).
For example,
$$A'P+OP = r,$$
where $r$ is the circle radius, and, by triangular inequality on triangle $AOP$,
$$AP+OP geq r,$$
so that
$$AP geq A'P.$$
So no matter where you take $A$ the distance from $P$ will be greater than that of $A'$ from $P$. Thus
$$A'P=2 mbox{cm}.$$
Similarly, you can express $A''P$ as
$$A''P = r + OP,$$
whereas, again for triangular inequality on $AOP$,
$$AP leq r + OP,$$
which yields
$$ AP leq A''P.$$
In conclusion $A''P = 8$ cm and the radius is $r = frac{A'P + A''P}{2}=5$ cm.
$hskip1.5in$
answered Jan 24 at 11:43
MatteoMatteo
888312
$endgroup$
You can use triangular inequality to show that the minimum and maximum distances are attained when the point $A$ is on the diameter passing through $P$ (see Figure in the bottom).
For example,
$$A'P+OP = r,$$
where $r$ is the circle radius, and, by triangular inequality on triangle $AOP$,
$$AP+OP geq r,$$
so that
$$AP geq A'P.$$
So no matter where you take $A$ the distance from $P$ will be greater than that of $A'$ from $P$. Thus
$$A'P=2 mbox{cm}.$$
Similarly, you can express $A''P$ as
$$A''P = r + OP,$$
whereas, again for triangular inequality on $AOP$,
$$AP leq r + OP,$$
which yields
$$ AP leq A''P.$$
In conclusion $A''P = 8$ cm and the radius is $r = frac{A'P + A''P}{2}=5$ cm.
$hskip1.5in$
answered Jan 24 at 11:43
MatteoMatteo
888312
edited Jan 24 at 19:09
answered Jan 24 at 11:43
MatteoMatteo
888312
answered Jan 24 at 11:43
MatteoMatteo
888312
answered Jan 24 at 11:43
MatteoMatteo
888312
888312
$begingroup$
Ok. then while doing same with A'',, will i take triangle AOA'' ?? Is AP=2 cm ??
$endgroup$
– Subham Sen
Jan 24 at 12:22
$begingroup$
@SubhamSen: $AP$ is greater than $A'P$, as we just demonstrated, right? Since $2$ cm is the minimum distance, then which segment is $2$ cm long? For the other question: try expressing $A''P$ in terms of the radius.
$endgroup$
– Matteo
Jan 24 at 13:04
$begingroup$
@SubhamSen added the conclusion in the text.
$endgroup$
– Matteo
Jan 24 at 14:21
$begingroup$
a> I couldn't get how A'P=2 cm and A''P=8 cm ??
$endgroup$
– Subham Sen
Jan 24 at 18:53
$begingroup$
b> Why are you saying, " no matter where you take A the distance from P will be greater than that of A' " ??
$endgroup$
– Subham Sen
Jan 24 at 18:56
|
show 2 more comments
$begingroup$
Ok. then while doing same with A'',, will i take triangle AOA'' ?? Is AP=2 cm ??
$endgroup$
– Subham Sen
Jan 24 at 12:22
$begingroup$
@SubhamSen: $AP$ is greater than $A'P$, as we just demonstrated, right? Since $2$ cm is the minimum distance, then which segment is $2$ cm long? For the other question: try expressing $A''P$ in terms of the radius.
$endgroup$
– Matteo
Jan 24 at 13:04
$begingroup$
@SubhamSen added the conclusion in the text.
$endgroup$
– Matteo
Jan 24 at 14:21
$begingroup$
a> I couldn't get how A'P=2 cm and A''P=8 cm ??
$endgroup$
– Subham Sen
Jan 24 at 18:53
$begingroup$
b> Why are you saying, " no matter where you take A the distance from P will be greater than that of A' " ??
$endgroup$
– Subham Sen
Jan 24 at 18:56
$begingroup$
Ok. then while doing same with A'',, will i take triangle AOA'' ?? Is AP=2 cm ??
$endgroup$
– Subham Sen
Jan 24 at 12:22
$begingroup$
Ok. then while doing same with A'',, will i take triangle AOA'' ?? Is AP=2 cm ??
$endgroup$
– Subham Sen
Jan 24 at 12:22
$begingroup$
@SubhamSen: $AP$ is greater than $A'P$, as we just demonstrated, right? Since $2$ cm is the minimum distance, then which segment is $2$ cm long? For the other question: try expressing $A''P$ in terms of the radius.
$endgroup$
– Matteo
Jan 24 at 13:04
$begingroup$
@SubhamSen: $AP$ is greater than $A'P$, as we just demonstrated, right? Since $2$ cm is the minimum distance, then which segment is $2$ cm long? For the other question: try expressing $A''P$ in terms of the radius.
$endgroup$
– Matteo
Jan 24 at 13:04
$begingroup$
@SubhamSen added the conclusion in the text.
$endgroup$
– Matteo
Jan 24 at 14:21
$begingroup$
@SubhamSen added the conclusion in the text.
$endgroup$
– Matteo
Jan 24 at 14:21
$begingroup$
a> I couldn't get how A'P=2 cm and A''P=8 cm ??
$endgroup$
– Subham Sen
Jan 24 at 18:53
$begingroup$
a> I couldn't get how A'P=2 cm and A''P=8 cm ??
$endgroup$
– Subham Sen
Jan 24 at 18:53
$begingroup$
b> Why are you saying, " no matter where you take A the distance from P will be greater than that of A' " ??
$endgroup$
– Subham Sen
Jan 24 at 18:56
$begingroup$
b> Why are you saying, " no matter where you take A the distance from P will be greater than that of A' " ??
$endgroup$
– Subham Sen
Jan 24 at 18:56
|
show 2 more comments
$begingroup$
The minimum distance from the point on the circumference will be at a right angle to the edge of the circle and the maximum distance will pass through the center point and also be at a right angle. Adding to two distances will make the diameter of the circle: $8+2 = 10$cm, therefore the radius is $5$ cm.
edited Jan 24 at 14:08
Namaste
1
answered Jan 24 at 13:56
JamesJames
1
$endgroup$
add a comment |
$begingroup$
The minimum distance from the point on the circumference will be at a right angle to the edge of the circle and the maximum distance will pass through the center point and also be at a right angle. Adding to two distances will make the diameter of the circle: $8+2 = 10$cm, therefore the radius is $5$ cm.
edited Jan 24 at 14:08
Namaste
1
answered Jan 24 at 13:56
JamesJames
1
$endgroup$
add a comment |
$begingroup$
The minimum distance from the point on the circumference will be at a right angle to the edge of the circle and the maximum distance will pass through the center point and also be at a right angle. Adding to two distances will make the diameter of the circle: $8+2 = 10$cm, therefore the radius is $5$ cm.
edited Jan 24 at 14:08
Namaste
1
answered Jan 24 at 13:56
JamesJames
1
$endgroup$
The minimum distance from the point on the circumference will be at a right angle to the edge of the circle and the maximum distance will pass through the center point and also be at a right angle. Adding to two distances will make the diameter of the circle: $8+2 = 10$cm, therefore the radius is $5$ cm.
edited Jan 24 at 14:08
Namaste
1
answered Jan 24 at 13:56
JamesJames
1
edited Jan 24 at 14:08
Namaste
1
edited Jan 24 at 14:08
Namaste
1
edited Jan 24 at 14:08
Namaste
1
1
answered Jan 24 at 13:56
JamesJames
1
answered Jan 24 at 13:56
JamesJames
1
answered Jan 24 at 13:56
JamesJames
1
1
add a comment |
add a comment |
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$begingroup$
Is there a rule as to how $P$ is chosen other than it being inside the circle? If not, shouldn't the minimum distance between $A$ and $P$ be $0$? What makes you think the radius is $>4$?
$endgroup$
– Shubham Johri
Jan 24 at 11:09
$begingroup$
Are you familiar with triangular inequality?
$endgroup$
– Matteo
Jan 24 at 11:16
$begingroup$
@ShubhamJohri Minimum AP=2 cm is given in the question. Since P is inside a circle and maximum AP=8 cm, so the diameter is greater than 8 cm and therefore, radius > 4 cm.
$endgroup$
– Subham Sen
Jan 24 at 12:33