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
geometry trigonometry
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add a comment |
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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
geometry trigonometry
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If you have the equation in Cartesian coordinates, can't you put $x = rcostheta$ and $y=rsintheta$ to get the equation in polar coordinates?
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– Minus One-Twelfth
Mar 20 at 0:57
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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.
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– amd
Mar 20 at 1:19
add a comment |
$begingroup$
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
geometry trigonometry
$endgroup$
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
geometry trigonometry
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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