Find the equation of the hyperbolic line through $A=(3,4)$, perpendicular to hyperbolic line $x^2+y^2=25$, $...
0
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I don't know how to solve this exercise in Hyperbolic Geometry.
Find the linear equation of hyperbolic line which passes through point $A=(3,4)$ and it is perpendicular to hyperbolic line with linear equation $x^2+y^2=25$ , $ y>0$.
hyperbolic-geometry
edited Dec 15 '18 at 17:05
Blue
47.8k870152
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
$endgroup$
add a comment |
0
$begingroup$
I don't know how to solve this exercise in Hyperbolic Geometry.
Find the linear equation of hyperbolic line which passes through point $A=(3,4)$ and it is perpendicular to hyperbolic line with linear equation $x^2+y^2=25$ , $ y>0$.
hyperbolic-geometry
edited Dec 15 '18 at 17:05
Blue
47.8k870152
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
$endgroup$
$begingroup$
To be clear: Is this an exercise in the Upper Half-Plane Model?
$endgroup$
– Blue
Dec 15 '18 at 17:07
$begingroup$
@Blue yes it is
$endgroup$
– Sotiris Z.
Dec 15 '18 at 17:10
$begingroup$
@Blue Can you help me?
$endgroup$
– Sotiris Z.
Dec 18 '18 at 20:30
$begingroup$
Since $A$ is on the given circle, the perpendicular "line" needed is the orthogonal circle through $A$. A neat thing about orthogonal circles: Where they meet, the tangent line to one circle passes through the center of the other. And, here, you know that the center of the target circle is on the $x$-axis. So, figure out where the tangent at $A$ meets the $x$-axis (recall that the tangent at a point is perpendicular to the radius to the point), and you'll have the center of the target circle; the distance from $A$ to that center gives you the radius; from there you can get the equation.
$endgroup$
– Blue
Dec 19 '18 at 2:53
add a comment |
0
0
0
$begingroup$
I don't know how to solve this exercise in Hyperbolic Geometry.
Find the linear equation of hyperbolic line which passes through point $A=(3,4)$ and it is perpendicular to hyperbolic line with linear equation $x^2+y^2=25$ , $ y>0$.
hyperbolic-geometry
edited Dec 15 '18 at 17:05
Blue
47.8k870152
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
$endgroup$
I don't know how to solve this exercise in Hyperbolic Geometry.
Find the linear equation of hyperbolic line which passes through point $A=(3,4)$ and it is perpendicular to hyperbolic line with linear equation $x^2+y^2=25$ , $ y>0$.
hyperbolic-geometry
hyperbolic-geometry
edited Dec 15 '18 at 17:05
Blue
47.8k870152
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
edited Dec 15 '18 at 17:05
Blue
47.8k870152
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
edited Dec 15 '18 at 17:05
Blue
47.8k870152
edited Dec 15 '18 at 17:05
Blue
47.8k870152
edited Dec 15 '18 at 17:05
Blue
47.8k870152
47.8k870152
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
125
$begingroup$
To be clear: Is this an exercise in the Upper Half-Plane Model?
$endgroup$
– Blue
Dec 15 '18 at 17:07
$begingroup$
@Blue yes it is
$endgroup$
– Sotiris Z.
Dec 15 '18 at 17:10
$begingroup$
@Blue Can you help me?
$endgroup$
– Sotiris Z.
Dec 18 '18 at 20:30
$begingroup$
Since $A$ is on the given circle, the perpendicular "line" needed is the orthogonal circle through $A$. A neat thing about orthogonal circles: Where they meet, the tangent line to one circle passes through the center of the other. And, here, you know that the center of the target circle is on the $x$-axis. So, figure out where the tangent at $A$ meets the $x$-axis (recall that the tangent at a point is perpendicular to the radius to the point), and you'll have the center of the target circle; the distance from $A$ to that center gives you the radius; from there you can get the equation.
$endgroup$
– Blue
Dec 19 '18 at 2:53
add a comment |
$begingroup$
To be clear: Is this an exercise in the Upper Half-Plane Model?
$endgroup$
– Blue
Dec 15 '18 at 17:07
$begingroup$
@Blue yes it is
$endgroup$
– Sotiris Z.
Dec 15 '18 at 17:10
$begingroup$
@Blue Can you help me?
$endgroup$
– Sotiris Z.
Dec 18 '18 at 20:30
$begingroup$
Since $A$ is on the given circle, the perpendicular "line" needed is the orthogonal circle through $A$. A neat thing about orthogonal circles: Where they meet, the tangent line to one circle passes through the center of the other. And, here, you know that the center of the target circle is on the $x$-axis. So, figure out where the tangent at $A$ meets the $x$-axis (recall that the tangent at a point is perpendicular to the radius to the point), and you'll have the center of the target circle; the distance from $A$ to that center gives you the radius; from there you can get the equation.
$endgroup$
– Blue
Dec 19 '18 at 2:53
$begingroup$
To be clear: Is this an exercise in the Upper Half-Plane Model?
$endgroup$
– Blue
Dec 15 '18 at 17:07
$begingroup$
To be clear: Is this an exercise in the Upper Half-Plane Model?
$endgroup$
– Blue
Dec 15 '18 at 17:07
$begingroup$
@Blue yes it is
$endgroup$
– Sotiris Z.
Dec 15 '18 at 17:10
$begingroup$
@Blue yes it is
$endgroup$
– Sotiris Z.
Dec 15 '18 at 17:10
$begingroup$
@Blue Can you help me?
$endgroup$
– Sotiris Z.
Dec 18 '18 at 20:30
$begingroup$
@Blue Can you help me?
$endgroup$
– Sotiris Z.
Dec 18 '18 at 20:30
$begingroup$
Since $A$ is on the given circle, the perpendicular "line" needed is the orthogonal circle through $A$. A neat thing about orthogonal circles: Where they meet, the tangent line to one circle passes through the center of the other. And, here, you know that the center of the target circle is on the $x$-axis. So, figure out where the tangent at $A$ meets the $x$-axis (recall that the tangent at a point is perpendicular to the radius to the point), and you'll have the center of the target circle; the distance from $A$ to that center gives you the radius; from there you can get the equation.
$endgroup$
– Blue
Dec 19 '18 at 2:53
$begingroup$
Since $A$ is on the given circle, the perpendicular "line" needed is the orthogonal circle through $A$. A neat thing about orthogonal circles: Where they meet, the tangent line to one circle passes through the center of the other. And, here, you know that the center of the target circle is on the $x$-axis. So, figure out where the tangent at $A$ meets the $x$-axis (recall that the tangent at a point is perpendicular to the radius to the point), and you'll have the center of the target circle; the distance from $A$ to that center gives you the radius; from there you can get the equation.
$endgroup$
– Blue
Dec 19 '18 at 2:53
add a comment |
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$begingroup$
To be clear: Is this an exercise in the Upper Half-Plane Model?
$endgroup$
– Blue
Dec 15 '18 at 17:07
$begingroup$
@Blue yes it is
$endgroup$
– Sotiris Z.
Dec 15 '18 at 17:10
$begingroup$
@Blue Can you help me?
$endgroup$
– Sotiris Z.
Dec 18 '18 at 20:30
$begingroup$
Since $A$ is on the given circle, the perpendicular "line" needed is the orthogonal circle through $A$. A neat thing about orthogonal circles: Where they meet, the tangent line to one circle passes through the center of the other. And, here, you know that the center of the target circle is on the $x$-axis. So, figure out where the tangent at $A$ meets the $x$-axis (recall that the tangent at a point is perpendicular to the radius to the point), and you'll have the center of the target circle; the distance from $A$ to that center gives you the radius; from there you can get the equation.
$endgroup$
– Blue
Dec 19 '18 at 2:53