Chord of contact in polar coordinateCoordinates of Intersection of two circlesTo prove that the centre of 2...

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Chord of contact in polar coordinate


Coordinates of Intersection of two circlesTo prove that the centre of 2 circles and the two points at which the 2 circles cut and the origin lie on a circle.Plotting polar equations of circles not centered at (0, 0)Identify the locus.Can we find a point $M$ on the unit circle such that $prod_{i=1}^n MA_i=1$?Tangent of a circleConics and Loci Question (Hyperbolae and Circles)Find radius based on length chord of a segment of a circleHow to Calculate Radius of Circle Given Two Points and Tangential CircleA Problem With Coordinate Systems













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Can you please help me to derive the equation of chord of contact for a circle in polar coordinate? I have found the equation in cartesian coordinate but I cannot map that into the polar coordinate. Any help would be appreciated.



Let suppose we want to derive the chord of contact that connects two points, e.g., $A(-pi/3,1)$ and $B(pi/4,1)$ on the unit circle. Center of the circle is $(0,0)$.



All the best,

Russell










share|cite|improve this question











$endgroup$












  • $begingroup$
    If you have the equation in Cartesian coordinates, can't you put $x = rcostheta$ and $y=rsintheta$ to get the equation in polar coordinates?
    $endgroup$
    – Minus One-Twelfth
    Mar 20 at 0:57










  • $begingroup$
    It’s not very hard to determine the distance of the line from the origin and its direction directly in polar coordinates. From there, it’s a simple matter of plugging the numbers into one of the two standard polar equations of lines.
    $endgroup$
    – amd
    Mar 20 at 1:19
















0












$begingroup$


Can you please help me to derive the equation of chord of contact for a circle in polar coordinate? I have found the equation in cartesian coordinate but I cannot map that into the polar coordinate. Any help would be appreciated.



Let suppose we want to derive the chord of contact that connects two points, e.g., $A(-pi/3,1)$ and $B(pi/4,1)$ on the unit circle. Center of the circle is $(0,0)$.



All the best,

Russell










share|cite|improve this question











$endgroup$












  • $begingroup$
    If you have the equation in Cartesian coordinates, can't you put $x = rcostheta$ and $y=rsintheta$ to get the equation in polar coordinates?
    $endgroup$
    – Minus One-Twelfth
    Mar 20 at 0:57










  • $begingroup$
    It’s not very hard to determine the distance of the line from the origin and its direction directly in polar coordinates. From there, it’s a simple matter of plugging the numbers into one of the two standard polar equations of lines.
    $endgroup$
    – amd
    Mar 20 at 1:19














0












0








0





$begingroup$


Can you please help me to derive the equation of chord of contact for a circle in polar coordinate? I have found the equation in cartesian coordinate but I cannot map that into the polar coordinate. Any help would be appreciated.



Let suppose we want to derive the chord of contact that connects two points, e.g., $A(-pi/3,1)$ and $B(pi/4,1)$ on the unit circle. Center of the circle is $(0,0)$.



All the best,

Russell










share|cite|improve this question











$endgroup$




Can you please help me to derive the equation of chord of contact for a circle in polar coordinate? I have found the equation in cartesian coordinate but I cannot map that into the polar coordinate. Any help would be appreciated.



Let suppose we want to derive the chord of contact that connects two points, e.g., $A(-pi/3,1)$ and $B(pi/4,1)$ on the unit circle. Center of the circle is $(0,0)$.



All the best,

Russell







geometry trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 20 at 1:22









Saad

20.4k92352




20.4k92352










asked Mar 20 at 0:54









NarimaniNarimani

1




1












  • $begingroup$
    If you have the equation in Cartesian coordinates, can't you put $x = rcostheta$ and $y=rsintheta$ to get the equation in polar coordinates?
    $endgroup$
    – Minus One-Twelfth
    Mar 20 at 0:57










  • $begingroup$
    It’s not very hard to determine the distance of the line from the origin and its direction directly in polar coordinates. From there, it’s a simple matter of plugging the numbers into one of the two standard polar equations of lines.
    $endgroup$
    – amd
    Mar 20 at 1:19


















  • $begingroup$
    If you have the equation in Cartesian coordinates, can't you put $x = rcostheta$ and $y=rsintheta$ to get the equation in polar coordinates?
    $endgroup$
    – Minus One-Twelfth
    Mar 20 at 0:57










  • $begingroup$
    It’s not very hard to determine the distance of the line from the origin and its direction directly in polar coordinates. From there, it’s a simple matter of plugging the numbers into one of the two standard polar equations of lines.
    $endgroup$
    – amd
    Mar 20 at 1:19
















$begingroup$
If you have the equation in Cartesian coordinates, can't you put $x = rcostheta$ and $y=rsintheta$ to get the equation in polar coordinates?
$endgroup$
– Minus One-Twelfth
Mar 20 at 0:57




$begingroup$
If you have the equation in Cartesian coordinates, can't you put $x = rcostheta$ and $y=rsintheta$ to get the equation in polar coordinates?
$endgroup$
– Minus One-Twelfth
Mar 20 at 0:57












$begingroup$
It’s not very hard to determine the distance of the line from the origin and its direction directly in polar coordinates. From there, it’s a simple matter of plugging the numbers into one of the two standard polar equations of lines.
$endgroup$
– amd
Mar 20 at 1:19




$begingroup$
It’s not very hard to determine the distance of the line from the origin and its direction directly in polar coordinates. From there, it’s a simple matter of plugging the numbers into one of the two standard polar equations of lines.
$endgroup$
– amd
Mar 20 at 1:19










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