Find the equation of the circle whose diameter is a chord.












4















$Y=mx$ is a chord of circle of radius $a$ through the origin whose
diameter is along the $x$-axis. Find the equation of the circle whose
diameter is the chord.




We also need to find the locus of its centre. I got a relation $h=m^2 h+(a^2+c)^.5$. Where $h$ is abscissa of the centre, $c$ is the constant term in the circle's equation.






circle







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    Would you please format mathematics as per the rules?
    – dbanet
    Aug 13 '16 at 9:13
















4















$Y=mx$ is a chord of circle of radius $a$ through the origin whose
diameter is along the $x$-axis. Find the equation of the circle whose
diameter is the chord.




We also need to find the locus of its centre. I got a relation $h=m^2 h+(a^2+c)^.5$. Where $h$ is abscissa of the centre, $c$ is the constant term in the circle's equation.






circle







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  • 1




    Would you please format mathematics as per the rules?
    – dbanet
    Aug 13 '16 at 9:13














4












4








4


1






$Y=mx$ is a chord of circle of radius $a$ through the origin whose
diameter is along the $x$-axis. Find the equation of the circle whose
diameter is the chord.




We also need to find the locus of its centre. I got a relation $h=m^2 h+(a^2+c)^.5$. Where $h$ is abscissa of the centre, $c$ is the constant term in the circle's equation.






circle







share|cite|improve this question
















$Y=mx$ is a chord of circle of radius $a$ through the origin whose
diameter is along the $x$-axis. Find the equation of the circle whose
diameter is the chord.




We also need to find the locus of its centre. I got a relation $h=m^2 h+(a^2+c)^.5$. Where $h$ is abscissa of the centre, $c$ is the constant term in the circle's equation.






circle




circle






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share|cite|improve this question








edited Aug 13 '16 at 9:40









PHPirate

328417




328417










asked Aug 13 '16 at 9:11









Areesh-Ul Eimaan

586




586








  • 1




    Would you please format mathematics as per the rules?
    – dbanet
    Aug 13 '16 at 9:13














  • 1




    Would you please format mathematics as per the rules?
    – dbanet
    Aug 13 '16 at 9:13








1




1




Would you please format mathematics as per the rules?
– dbanet
Aug 13 '16 at 9:13




Would you please format mathematics as per the rules?
– dbanet
Aug 13 '16 at 9:13










2 Answers
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The centre of the given circle lies at the point
$$(a,0)$$
because it passes through the origin and its diameter is along the x-axis.
The equation of this circle can be written as:
$$(x-a)^2+(y-0)^2 = a^2
\x^2+y^2-2ax = 0$$
The equation of the family of circles passing through the points of intersection of a circle $$S=0$$ and a line $$L=0$$ is given by:
$$S+lambda{L} = 0$$
The equation of the family of circles passing through $$x^2+y^2-2ax = 0$$ and $$y-mx = 0$$ is given by:
$$x^2+y^2-2ax+lambda{(y-mx)} = 0
\x^2+y^2-(2a+lambda{m})x+lambda{y} = 0$$
Coordinates of the centre of a circle are given by
$$(-g,-f)$$
where 2g and 2f are respectively the x-coefficient and y-coefficient in the equation of the circle.
Hence, the coordinates of the centre, C of the family of circles are given by:
$$left(frac{2a+lambda{m}}{2},frac{-lambda}{2}right)$$
The centre of the required circle lies on the line $$y = mx$$ because it is the diameter of the circle. Hence the coordinates of point C must satisfy the equation of this line.
$$y = mx
\frac{-lambda}{2} = mleft(frac{2a+lambda{m}}{2}right)$$
On solving, we get
$$lambda = frac{-2am}{1+m^2}$$
On putting $$lambda = frac{-2am}{1+m^2}$$ in the equation of the family of circles, we get the equation of the required circle as:
$$(1+m^2)(x^2+y^2)-2a(x+my) = 0$$






share|cite|improve this answer































    0














    As the diameter of the circle is on the x axis.so it's center doesn't have any $y$ co-ordinate.and as it's radius is $a$,we get the equation of the circle is-
    $$ (x-a)^2+y^2=a^2...........(1)$$
    the equation of the chord is $$y=mx.........(2)$$
    solving equation (1) and (2) we get, $$(x,y)=(0,0) and bigg(dfrac{2a}{m^2+1},dfrac{2am}{m^2+1}bigg)$$
    So ,the locus of the center of our required circle is $bigg(dfrac{a}{m^2+1},dfrac{am}{m^2+1}bigg)$ and the radius of the circle is $$dfrac{sqrt{bigg(dfrac{2a}{m^2+1}-0bigg)^2+bigg(dfrac{2am}{m^2+1}-0bigg)^2}}{2}=dfrac{a}{sqrt{m^2+1}}$$
    Hence We can now make our required equation and that is-
    $$bigg(x-dfrac{a}{m^2+1}bigg)^2+bigg(y-dfrac{am}{m^2+1}bigg)^2=dfrac{a^2}{m^2+1}$$
    After some simplification you will find that the equation become,
    $$2a(my+x)=(m^2+1)(x^2+y^2)$$
    graph of the circle






    share|cite|improve this answer























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      2 Answers
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      2 Answers
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      The centre of the given circle lies at the point
      $$(a,0)$$
      because it passes through the origin and its diameter is along the x-axis.
      The equation of this circle can be written as:
      $$(x-a)^2+(y-0)^2 = a^2
      \x^2+y^2-2ax = 0$$
      The equation of the family of circles passing through the points of intersection of a circle $$S=0$$ and a line $$L=0$$ is given by:
      $$S+lambda{L} = 0$$
      The equation of the family of circles passing through $$x^2+y^2-2ax = 0$$ and $$y-mx = 0$$ is given by:
      $$x^2+y^2-2ax+lambda{(y-mx)} = 0
      \x^2+y^2-(2a+lambda{m})x+lambda{y} = 0$$
      Coordinates of the centre of a circle are given by
      $$(-g,-f)$$
      where 2g and 2f are respectively the x-coefficient and y-coefficient in the equation of the circle.
      Hence, the coordinates of the centre, C of the family of circles are given by:
      $$left(frac{2a+lambda{m}}{2},frac{-lambda}{2}right)$$
      The centre of the required circle lies on the line $$y = mx$$ because it is the diameter of the circle. Hence the coordinates of point C must satisfy the equation of this line.
      $$y = mx
      \frac{-lambda}{2} = mleft(frac{2a+lambda{m}}{2}right)$$
      On solving, we get
      $$lambda = frac{-2am}{1+m^2}$$
      On putting $$lambda = frac{-2am}{1+m^2}$$ in the equation of the family of circles, we get the equation of the required circle as:
      $$(1+m^2)(x^2+y^2)-2a(x+my) = 0$$






      share|cite|improve this answer




























        0














        The centre of the given circle lies at the point
        $$(a,0)$$
        because it passes through the origin and its diameter is along the x-axis.
        The equation of this circle can be written as:
        $$(x-a)^2+(y-0)^2 = a^2
        \x^2+y^2-2ax = 0$$
        The equation of the family of circles passing through the points of intersection of a circle $$S=0$$ and a line $$L=0$$ is given by:
        $$S+lambda{L} = 0$$
        The equation of the family of circles passing through $$x^2+y^2-2ax = 0$$ and $$y-mx = 0$$ is given by:
        $$x^2+y^2-2ax+lambda{(y-mx)} = 0
        \x^2+y^2-(2a+lambda{m})x+lambda{y} = 0$$
        Coordinates of the centre of a circle are given by
        $$(-g,-f)$$
        where 2g and 2f are respectively the x-coefficient and y-coefficient in the equation of the circle.
        Hence, the coordinates of the centre, C of the family of circles are given by:
        $$left(frac{2a+lambda{m}}{2},frac{-lambda}{2}right)$$
        The centre of the required circle lies on the line $$y = mx$$ because it is the diameter of the circle. Hence the coordinates of point C must satisfy the equation of this line.
        $$y = mx
        \frac{-lambda}{2} = mleft(frac{2a+lambda{m}}{2}right)$$
        On solving, we get
        $$lambda = frac{-2am}{1+m^2}$$
        On putting $$lambda = frac{-2am}{1+m^2}$$ in the equation of the family of circles, we get the equation of the required circle as:
        $$(1+m^2)(x^2+y^2)-2a(x+my) = 0$$






        share|cite|improve this answer


























          0












          0








          0






          The centre of the given circle lies at the point
          $$(a,0)$$
          because it passes through the origin and its diameter is along the x-axis.
          The equation of this circle can be written as:
          $$(x-a)^2+(y-0)^2 = a^2
          \x^2+y^2-2ax = 0$$
          The equation of the family of circles passing through the points of intersection of a circle $$S=0$$ and a line $$L=0$$ is given by:
          $$S+lambda{L} = 0$$
          The equation of the family of circles passing through $$x^2+y^2-2ax = 0$$ and $$y-mx = 0$$ is given by:
          $$x^2+y^2-2ax+lambda{(y-mx)} = 0
          \x^2+y^2-(2a+lambda{m})x+lambda{y} = 0$$
          Coordinates of the centre of a circle are given by
          $$(-g,-f)$$
          where 2g and 2f are respectively the x-coefficient and y-coefficient in the equation of the circle.
          Hence, the coordinates of the centre, C of the family of circles are given by:
          $$left(frac{2a+lambda{m}}{2},frac{-lambda}{2}right)$$
          The centre of the required circle lies on the line $$y = mx$$ because it is the diameter of the circle. Hence the coordinates of point C must satisfy the equation of this line.
          $$y = mx
          \frac{-lambda}{2} = mleft(frac{2a+lambda{m}}{2}right)$$
          On solving, we get
          $$lambda = frac{-2am}{1+m^2}$$
          On putting $$lambda = frac{-2am}{1+m^2}$$ in the equation of the family of circles, we get the equation of the required circle as:
          $$(1+m^2)(x^2+y^2)-2a(x+my) = 0$$






          share|cite|improve this answer














          The centre of the given circle lies at the point
          $$(a,0)$$
          because it passes through the origin and its diameter is along the x-axis.
          The equation of this circle can be written as:
          $$(x-a)^2+(y-0)^2 = a^2
          \x^2+y^2-2ax = 0$$
          The equation of the family of circles passing through the points of intersection of a circle $$S=0$$ and a line $$L=0$$ is given by:
          $$S+lambda{L} = 0$$
          The equation of the family of circles passing through $$x^2+y^2-2ax = 0$$ and $$y-mx = 0$$ is given by:
          $$x^2+y^2-2ax+lambda{(y-mx)} = 0
          \x^2+y^2-(2a+lambda{m})x+lambda{y} = 0$$
          Coordinates of the centre of a circle are given by
          $$(-g,-f)$$
          where 2g and 2f are respectively the x-coefficient and y-coefficient in the equation of the circle.
          Hence, the coordinates of the centre, C of the family of circles are given by:
          $$left(frac{2a+lambda{m}}{2},frac{-lambda}{2}right)$$
          The centre of the required circle lies on the line $$y = mx$$ because it is the diameter of the circle. Hence the coordinates of point C must satisfy the equation of this line.
          $$y = mx
          \frac{-lambda}{2} = mleft(frac{2a+lambda{m}}{2}right)$$
          On solving, we get
          $$lambda = frac{-2am}{1+m^2}$$
          On putting $$lambda = frac{-2am}{1+m^2}$$ in the equation of the family of circles, we get the equation of the required circle as:
          $$(1+m^2)(x^2+y^2)-2a(x+my) = 0$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jun 26 '17 at 22:11

























          answered Jun 26 '17 at 21:59









          Himanshu

          887




          887























              0














              As the diameter of the circle is on the x axis.so it's center doesn't have any $y$ co-ordinate.and as it's radius is $a$,we get the equation of the circle is-
              $$ (x-a)^2+y^2=a^2...........(1)$$
              the equation of the chord is $$y=mx.........(2)$$
              solving equation (1) and (2) we get, $$(x,y)=(0,0) and bigg(dfrac{2a}{m^2+1},dfrac{2am}{m^2+1}bigg)$$
              So ,the locus of the center of our required circle is $bigg(dfrac{a}{m^2+1},dfrac{am}{m^2+1}bigg)$ and the radius of the circle is $$dfrac{sqrt{bigg(dfrac{2a}{m^2+1}-0bigg)^2+bigg(dfrac{2am}{m^2+1}-0bigg)^2}}{2}=dfrac{a}{sqrt{m^2+1}}$$
              Hence We can now make our required equation and that is-
              $$bigg(x-dfrac{a}{m^2+1}bigg)^2+bigg(y-dfrac{am}{m^2+1}bigg)^2=dfrac{a^2}{m^2+1}$$
              After some simplification you will find that the equation become,
              $$2a(my+x)=(m^2+1)(x^2+y^2)$$
              graph of the circle






              share|cite|improve this answer




























                0














                As the diameter of the circle is on the x axis.so it's center doesn't have any $y$ co-ordinate.and as it's radius is $a$,we get the equation of the circle is-
                $$ (x-a)^2+y^2=a^2...........(1)$$
                the equation of the chord is $$y=mx.........(2)$$
                solving equation (1) and (2) we get, $$(x,y)=(0,0) and bigg(dfrac{2a}{m^2+1},dfrac{2am}{m^2+1}bigg)$$
                So ,the locus of the center of our required circle is $bigg(dfrac{a}{m^2+1},dfrac{am}{m^2+1}bigg)$ and the radius of the circle is $$dfrac{sqrt{bigg(dfrac{2a}{m^2+1}-0bigg)^2+bigg(dfrac{2am}{m^2+1}-0bigg)^2}}{2}=dfrac{a}{sqrt{m^2+1}}$$
                Hence We can now make our required equation and that is-
                $$bigg(x-dfrac{a}{m^2+1}bigg)^2+bigg(y-dfrac{am}{m^2+1}bigg)^2=dfrac{a^2}{m^2+1}$$
                After some simplification you will find that the equation become,
                $$2a(my+x)=(m^2+1)(x^2+y^2)$$
                graph of the circle






                share|cite|improve this answer


























                  0












                  0








                  0






                  As the diameter of the circle is on the x axis.so it's center doesn't have any $y$ co-ordinate.and as it's radius is $a$,we get the equation of the circle is-
                  $$ (x-a)^2+y^2=a^2...........(1)$$
                  the equation of the chord is $$y=mx.........(2)$$
                  solving equation (1) and (2) we get, $$(x,y)=(0,0) and bigg(dfrac{2a}{m^2+1},dfrac{2am}{m^2+1}bigg)$$
                  So ,the locus of the center of our required circle is $bigg(dfrac{a}{m^2+1},dfrac{am}{m^2+1}bigg)$ and the radius of the circle is $$dfrac{sqrt{bigg(dfrac{2a}{m^2+1}-0bigg)^2+bigg(dfrac{2am}{m^2+1}-0bigg)^2}}{2}=dfrac{a}{sqrt{m^2+1}}$$
                  Hence We can now make our required equation and that is-
                  $$bigg(x-dfrac{a}{m^2+1}bigg)^2+bigg(y-dfrac{am}{m^2+1}bigg)^2=dfrac{a^2}{m^2+1}$$
                  After some simplification you will find that the equation become,
                  $$2a(my+x)=(m^2+1)(x^2+y^2)$$
                  graph of the circle






                  share|cite|improve this answer














                  As the diameter of the circle is on the x axis.so it's center doesn't have any $y$ co-ordinate.and as it's radius is $a$,we get the equation of the circle is-
                  $$ (x-a)^2+y^2=a^2...........(1)$$
                  the equation of the chord is $$y=mx.........(2)$$
                  solving equation (1) and (2) we get, $$(x,y)=(0,0) and bigg(dfrac{2a}{m^2+1},dfrac{2am}{m^2+1}bigg)$$
                  So ,the locus of the center of our required circle is $bigg(dfrac{a}{m^2+1},dfrac{am}{m^2+1}bigg)$ and the radius of the circle is $$dfrac{sqrt{bigg(dfrac{2a}{m^2+1}-0bigg)^2+bigg(dfrac{2am}{m^2+1}-0bigg)^2}}{2}=dfrac{a}{sqrt{m^2+1}}$$
                  Hence We can now make our required equation and that is-
                  $$bigg(x-dfrac{a}{m^2+1}bigg)^2+bigg(y-dfrac{am}{m^2+1}bigg)^2=dfrac{a^2}{m^2+1}$$
                  After some simplification you will find that the equation become,
                  $$2a(my+x)=(m^2+1)(x^2+y^2)$$
                  graph of the circle







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Sep 11 at 18:15

























                  answered Sep 11 at 18:09









                  Rakibul Islam Prince

                  880211




                  880211






























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