Co-ordinate Geometry : Circle












1












$begingroup$


Problem: Circles $x^2 + y^2 = 1$ and $x^2 + y^2 - 8x + 11 = 0$ cut off equal intercepts on a line through the point $(-2, frac{1}{2})$. Find the slope of the line.



Comments: First I write the family of lines passing through the given point i.e., $y - frac{1}{2} = m (x + 2)$ then by using the pythagorean theorem and the hypothesis I got a equation in $m$ i.e., $| 2m - 7/2| - |2m + 1/2| = 15 sqrt{1 + m^2}$ but the solution of this equation is lengthy.



I am seeking a shorter solution.






analytic-geometry circle







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  • $begingroup$
    Check case when $2m$ is between $-infty,-1/2,0; 0,7/2; 7/2,infty$
    $endgroup$
    – lab bhattacharjee
    Nov 30 '18 at 5:39
















1












$begingroup$


Problem: Circles $x^2 + y^2 = 1$ and $x^2 + y^2 - 8x + 11 = 0$ cut off equal intercepts on a line through the point $(-2, frac{1}{2})$. Find the slope of the line.



Comments: First I write the family of lines passing through the given point i.e., $y - frac{1}{2} = m (x + 2)$ then by using the pythagorean theorem and the hypothesis I got a equation in $m$ i.e., $| 2m - 7/2| - |2m + 1/2| = 15 sqrt{1 + m^2}$ but the solution of this equation is lengthy.



I am seeking a shorter solution.






analytic-geometry circle







share|cite|improve this question









$endgroup$












  • $begingroup$
    Check case when $2m$ is between $-infty,-1/2,0; 0,7/2; 7/2,infty$
    $endgroup$
    – lab bhattacharjee
    Nov 30 '18 at 5:39














1












1








1


2



$begingroup$


Problem: Circles $x^2 + y^2 = 1$ and $x^2 + y^2 - 8x + 11 = 0$ cut off equal intercepts on a line through the point $(-2, frac{1}{2})$. Find the slope of the line.



Comments: First I write the family of lines passing through the given point i.e., $y - frac{1}{2} = m (x + 2)$ then by using the pythagorean theorem and the hypothesis I got a equation in $m$ i.e., $| 2m - 7/2| - |2m + 1/2| = 15 sqrt{1 + m^2}$ but the solution of this equation is lengthy.



I am seeking a shorter solution.






analytic-geometry circle







share|cite|improve this question









$endgroup$




Problem: Circles $x^2 + y^2 = 1$ and $x^2 + y^2 - 8x + 11 = 0$ cut off equal intercepts on a line through the point $(-2, frac{1}{2})$. Find the slope of the line.



Comments: First I write the family of lines passing through the given point i.e., $y - frac{1}{2} = m (x + 2)$ then by using the pythagorean theorem and the hypothesis I got a equation in $m$ i.e., $| 2m - 7/2| - |2m + 1/2| = 15 sqrt{1 + m^2}$ but the solution of this equation is lengthy.



I am seeking a shorter solution.






analytic-geometry circle




analytic-geometry circle






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Nov 30 '18 at 4:35









prashant sharmaprashant sharma

757




757












  • $begingroup$
    Check case when $2m$ is between $-infty,-1/2,0; 0,7/2; 7/2,infty$
    $endgroup$
    – lab bhattacharjee
    Nov 30 '18 at 5:39


















  • $begingroup$
    Check case when $2m$ is between $-infty,-1/2,0; 0,7/2; 7/2,infty$
    $endgroup$
    – lab bhattacharjee
    Nov 30 '18 at 5:39
















$begingroup$
Check case when $2m$ is between $-infty,-1/2,0; 0,7/2; 7/2,infty$
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 5:39




$begingroup$
Check case when $2m$ is between $-infty,-1/2,0; 0,7/2; 7/2,infty$
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 5:39










1 Answer
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oldest

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4












$begingroup$


  • Line $$2mx-2y+4m+1=0 tag{1}$$



  • First circle $$x^2+y^2=1$$



    centre $O=(0,0)$, radius $r_1=1$



    distance of $O$ from $(1)$: $$d_1=left| frac{4m+1}{2sqrt{m^2+1}} right|$$




  • Second circle $$x^2+y^2-8x+11=0$$



    centre $P=(4,0)$, radius $r_2=sqrt{4^2-11}=sqrt{5}$



    distance of $P$ from $(1)$: $$d_2=left| frac{12m+1}{2sqrt{m^2+1}} right|$$




  • Equating semi-chord length: $r^2-d^2$



    begin{align}
    r_1^2-d_1^2 &= r_2^2-d_2^2 \
    1-frac{(4m+1)^2}{4(m^2+1)} &=
    5-frac{(12m+1)^2}{4(m^2+1)} \
    m &= frac{-1pm sqrt{29}}{14}
    end{align}




Only the negative root gives intersections:



enter image description here






share|cite|improve this answer











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    1 Answer
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    1






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    active

    oldest

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    4












    $begingroup$


    • Line $$2mx-2y+4m+1=0 tag{1}$$



    • First circle $$x^2+y^2=1$$



      centre $O=(0,0)$, radius $r_1=1$



      distance of $O$ from $(1)$: $$d_1=left| frac{4m+1}{2sqrt{m^2+1}} right|$$




    • Second circle $$x^2+y^2-8x+11=0$$



      centre $P=(4,0)$, radius $r_2=sqrt{4^2-11}=sqrt{5}$



      distance of $P$ from $(1)$: $$d_2=left| frac{12m+1}{2sqrt{m^2+1}} right|$$




    • Equating semi-chord length: $r^2-d^2$



      begin{align}
      r_1^2-d_1^2 &= r_2^2-d_2^2 \
      1-frac{(4m+1)^2}{4(m^2+1)} &=
      5-frac{(12m+1)^2}{4(m^2+1)} \
      m &= frac{-1pm sqrt{29}}{14}
      end{align}




    Only the negative root gives intersections:



    enter image description here






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$


      • Line $$2mx-2y+4m+1=0 tag{1}$$



      • First circle $$x^2+y^2=1$$



        centre $O=(0,0)$, radius $r_1=1$



        distance of $O$ from $(1)$: $$d_1=left| frac{4m+1}{2sqrt{m^2+1}} right|$$




      • Second circle $$x^2+y^2-8x+11=0$$



        centre $P=(4,0)$, radius $r_2=sqrt{4^2-11}=sqrt{5}$



        distance of $P$ from $(1)$: $$d_2=left| frac{12m+1}{2sqrt{m^2+1}} right|$$




      • Equating semi-chord length: $r^2-d^2$



        begin{align}
        r_1^2-d_1^2 &= r_2^2-d_2^2 \
        1-frac{(4m+1)^2}{4(m^2+1)} &=
        5-frac{(12m+1)^2}{4(m^2+1)} \
        m &= frac{-1pm sqrt{29}}{14}
        end{align}




      Only the negative root gives intersections:



      enter image description here






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$


        • Line $$2mx-2y+4m+1=0 tag{1}$$



        • First circle $$x^2+y^2=1$$



          centre $O=(0,0)$, radius $r_1=1$



          distance of $O$ from $(1)$: $$d_1=left| frac{4m+1}{2sqrt{m^2+1}} right|$$




        • Second circle $$x^2+y^2-8x+11=0$$



          centre $P=(4,0)$, radius $r_2=sqrt{4^2-11}=sqrt{5}$



          distance of $P$ from $(1)$: $$d_2=left| frac{12m+1}{2sqrt{m^2+1}} right|$$




        • Equating semi-chord length: $r^2-d^2$



          begin{align}
          r_1^2-d_1^2 &= r_2^2-d_2^2 \
          1-frac{(4m+1)^2}{4(m^2+1)} &=
          5-frac{(12m+1)^2}{4(m^2+1)} \
          m &= frac{-1pm sqrt{29}}{14}
          end{align}




        Only the negative root gives intersections:



        enter image description here






        share|cite|improve this answer











        $endgroup$




        • Line $$2mx-2y+4m+1=0 tag{1}$$



        • First circle $$x^2+y^2=1$$



          centre $O=(0,0)$, radius $r_1=1$



          distance of $O$ from $(1)$: $$d_1=left| frac{4m+1}{2sqrt{m^2+1}} right|$$




        • Second circle $$x^2+y^2-8x+11=0$$



          centre $P=(4,0)$, radius $r_2=sqrt{4^2-11}=sqrt{5}$



          distance of $P$ from $(1)$: $$d_2=left| frac{12m+1}{2sqrt{m^2+1}} right|$$




        • Equating semi-chord length: $r^2-d^2$



          begin{align}
          r_1^2-d_1^2 &= r_2^2-d_2^2 \
          1-frac{(4m+1)^2}{4(m^2+1)} &=
          5-frac{(12m+1)^2}{4(m^2+1)} \
          m &= frac{-1pm sqrt{29}}{14}
          end{align}




        Only the negative root gives intersections:



        enter image description here







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 2 '18 at 7:15

























        answered Nov 30 '18 at 6:46









        Ng Chung TakNg Chung Tak

        14.6k31334




        14.6k31334






























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