P is a point inside a circle and A is a point on the circumference.Find the radius of the circle.












0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0


0



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















0












$begingroup$

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.



$hskip1.5in$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















0












$begingroup$

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.






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    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.



    $hskip1.5in$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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
















    0












    $begingroup$

    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.



    $hskip1.5in$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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














    0












    0








    0





    $begingroup$

    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.



    $hskip1.5in$






    share|cite|improve this answer











    $endgroup$



    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.



    $hskip1.5in$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 24 at 19:09

























    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


















    • $begingroup$
      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











    0












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 24 at 14:08









        Namaste

        1




        1










        answered Jan 24 at 13:56









        JamesJames

        1




        1






























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