Equation of tangent on Cartesian plane given center and radius of a circle
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If I have a generic circle with radius $r$ and center $(h, k)$, and a tangent line with point of tangency $(x, y)$, can you give me the equation of the tangent line? Thanks!
geometry trigonometry circle coordinate-systems
edited May 31 '14 at 0:17
Cookie
8,720123682
asked May 31 '14 at 0:15
CarraraCarrara
323
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add a comment |
$begingroup$
If I have a generic circle with radius $r$ and center $(h, k)$, and a tangent line with point of tangency $(x, y)$, can you give me the equation of the tangent line? Thanks!
geometry trigonometry circle coordinate-systems
edited May 31 '14 at 0:17
Cookie
8,720123682
asked May 31 '14 at 0:15
CarraraCarrara
323
$endgroup$
add a comment |
$begingroup$
If I have a generic circle with radius $r$ and center $(h, k)$, and a tangent line with point of tangency $(x, y)$, can you give me the equation of the tangent line? Thanks!
geometry trigonometry circle coordinate-systems
edited May 31 '14 at 0:17
Cookie
8,720123682
asked May 31 '14 at 0:15
CarraraCarrara
323
$endgroup$
If I have a generic circle with radius $r$ and center $(h, k)$, and a tangent line with point of tangency $(x, y)$, can you give me the equation of the tangent line? Thanks!
geometry trigonometry circle coordinate-systems
geometry trigonometry circle coordinate-systems
edited May 31 '14 at 0:17
Cookie
8,720123682
asked May 31 '14 at 0:15
CarraraCarrara
323
edited May 31 '14 at 0:17
Cookie
8,720123682
asked May 31 '14 at 0:15
CarraraCarrara
323
edited May 31 '14 at 0:17
Cookie
8,720123682
edited May 31 '14 at 0:17
Cookie
8,720123682
edited May 31 '14 at 0:17
Cookie
8,720123682
8,720123682
asked May 31 '14 at 0:15
CarraraCarrara
323
asked May 31 '14 at 0:15
CarraraCarrara
323
asked May 31 '14 at 0:15
CarraraCarrara
323
323
add a comment |
add a comment |
2 Answers
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Here's how I was taught to get tangents to conic sections in high school, half a century ago. If you want to get the tangent at the point $(x_0,y_0)$, you take the equation of the conic, and replace "half" of the $x$'s by $x_0$ and "half" of the $y$'s by $y_0$.
In this case, the equation of the curve is $(x-h)^2+(y-k)^2=r^2$, so the tangent at $(x_0,y_0)$ will be $(x_0-h)(x-h) + (y_0-k)(y-k)=r^2$.
A little algebra, using the fact that $x_0^2 + y_0^2 = r^2$, shows that this can also be written in the form $(x_0 -h)(x-x_0) + (y_0 -k)(y-y_0) = 0$.
This works with any conic section curve, and it gives you tangent planes of quadric surfaces, too.
Of course, you can apply the same process even if the point $(x_0,y_0)$ doesn't lie on the curve. In this case, the line you get is called the polar line of the point. It just happens that, for any point lying on the curve, the polar line is the tangent at the point.
I'm curious -- do they still teach this stuff?
answered May 31 '14 at 4:10
bubbabubba
30.3k33086
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add a comment |
$begingroup$
Let the point of tangency be $P=(x_0, y_0)$. If the center of the circle is at $C=(h,k)$, then the line $overleftrightarrow{CP}$ is perpendicular to tangent line at the point $P$. The slope of the line $overleftrightarrow{CP}$ is $dfrac{y_0-k}{x_0-h}$, so the slope of the tangent line is $m=-dfrac{x_0-h}{y_0-k}$.
The equation of the tangent line is therefore
$$y - y_0 = -dfrac{x_0-h}{y_0-k}(x-x_0)$$
Just to satisfy my own curiosity, I want to write this equation is standard form.
begin{align}
y - y_0 &= -dfrac{x_0-h}{y_0-k}(x-x_0) \
(y_0-k)y - (y_0-k)y_0 &= -(x_0-h)x + (x_0-h)x_0 \
(x_0-h)x + (y_0-k)y &= (x_0-h)x_0 + (y_0-k)y_0 \
(x_0-h)x + (y_0-k)y &= x_0^2 + y_0^2 -h x_0 - k y_0\
end{align}
answered May 16 '18 at 23:24
steven gregorysteven gregory
18k32258
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2 Answers
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2 Answers
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active
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$begingroup$
Here's how I was taught to get tangents to conic sections in high school, half a century ago. If you want to get the tangent at the point $(x_0,y_0)$, you take the equation of the conic, and replace "half" of the $x$'s by $x_0$ and "half" of the $y$'s by $y_0$.
In this case, the equation of the curve is $(x-h)^2+(y-k)^2=r^2$, so the tangent at $(x_0,y_0)$ will be $(x_0-h)(x-h) + (y_0-k)(y-k)=r^2$.
A little algebra, using the fact that $x_0^2 + y_0^2 = r^2$, shows that this can also be written in the form $(x_0 -h)(x-x_0) + (y_0 -k)(y-y_0) = 0$.
This works with any conic section curve, and it gives you tangent planes of quadric surfaces, too.
Of course, you can apply the same process even if the point $(x_0,y_0)$ doesn't lie on the curve. In this case, the line you get is called the polar line of the point. It just happens that, for any point lying on the curve, the polar line is the tangent at the point.
I'm curious -- do they still teach this stuff?
answered May 31 '14 at 4:10
bubbabubba
30.3k33086
$endgroup$
add a comment |
$begingroup$
Here's how I was taught to get tangents to conic sections in high school, half a century ago. If you want to get the tangent at the point $(x_0,y_0)$, you take the equation of the conic, and replace "half" of the $x$'s by $x_0$ and "half" of the $y$'s by $y_0$.
In this case, the equation of the curve is $(x-h)^2+(y-k)^2=r^2$, so the tangent at $(x_0,y_0)$ will be $(x_0-h)(x-h) + (y_0-k)(y-k)=r^2$.
A little algebra, using the fact that $x_0^2 + y_0^2 = r^2$, shows that this can also be written in the form $(x_0 -h)(x-x_0) + (y_0 -k)(y-y_0) = 0$.
This works with any conic section curve, and it gives you tangent planes of quadric surfaces, too.
Of course, you can apply the same process even if the point $(x_0,y_0)$ doesn't lie on the curve. In this case, the line you get is called the polar line of the point. It just happens that, for any point lying on the curve, the polar line is the tangent at the point.
I'm curious -- do they still teach this stuff?
answered May 31 '14 at 4:10
bubbabubba
30.3k33086
$endgroup$
add a comment |
$begingroup$
Here's how I was taught to get tangents to conic sections in high school, half a century ago. If you want to get the tangent at the point $(x_0,y_0)$, you take the equation of the conic, and replace "half" of the $x$'s by $x_0$ and "half" of the $y$'s by $y_0$.
In this case, the equation of the curve is $(x-h)^2+(y-k)^2=r^2$, so the tangent at $(x_0,y_0)$ will be $(x_0-h)(x-h) + (y_0-k)(y-k)=r^2$.
A little algebra, using the fact that $x_0^2 + y_0^2 = r^2$, shows that this can also be written in the form $(x_0 -h)(x-x_0) + (y_0 -k)(y-y_0) = 0$.
This works with any conic section curve, and it gives you tangent planes of quadric surfaces, too.
Of course, you can apply the same process even if the point $(x_0,y_0)$ doesn't lie on the curve. In this case, the line you get is called the polar line of the point. It just happens that, for any point lying on the curve, the polar line is the tangent at the point.
I'm curious -- do they still teach this stuff?
answered May 31 '14 at 4:10
bubbabubba
30.3k33086
$endgroup$
Here's how I was taught to get tangents to conic sections in high school, half a century ago. If you want to get the tangent at the point $(x_0,y_0)$, you take the equation of the conic, and replace "half" of the $x$'s by $x_0$ and "half" of the $y$'s by $y_0$.
In this case, the equation of the curve is $(x-h)^2+(y-k)^2=r^2$, so the tangent at $(x_0,y_0)$ will be $(x_0-h)(x-h) + (y_0-k)(y-k)=r^2$.
A little algebra, using the fact that $x_0^2 + y_0^2 = r^2$, shows that this can also be written in the form $(x_0 -h)(x-x_0) + (y_0 -k)(y-y_0) = 0$.
This works with any conic section curve, and it gives you tangent planes of quadric surfaces, too.
Of course, you can apply the same process even if the point $(x_0,y_0)$ doesn't lie on the curve. In this case, the line you get is called the polar line of the point. It just happens that, for any point lying on the curve, the polar line is the tangent at the point.
I'm curious -- do they still teach this stuff?
answered May 31 '14 at 4:10
bubbabubba
30.3k33086
edited May 31 '14 at 6:54
answered May 31 '14 at 4:10
bubbabubba
30.3k33086
answered May 31 '14 at 4:10
bubbabubba
30.3k33086
answered May 31 '14 at 4:10
bubbabubba
30.3k33086
30.3k33086
add a comment |
add a comment |
$begingroup$
Let the point of tangency be $P=(x_0, y_0)$. If the center of the circle is at $C=(h,k)$, then the line $overleftrightarrow{CP}$ is perpendicular to tangent line at the point $P$. The slope of the line $overleftrightarrow{CP}$ is $dfrac{y_0-k}{x_0-h}$, so the slope of the tangent line is $m=-dfrac{x_0-h}{y_0-k}$.
The equation of the tangent line is therefore
$$y - y_0 = -dfrac{x_0-h}{y_0-k}(x-x_0)$$
Just to satisfy my own curiosity, I want to write this equation is standard form.
begin{align}
y - y_0 &= -dfrac{x_0-h}{y_0-k}(x-x_0) \
(y_0-k)y - (y_0-k)y_0 &= -(x_0-h)x + (x_0-h)x_0 \
(x_0-h)x + (y_0-k)y &= (x_0-h)x_0 + (y_0-k)y_0 \
(x_0-h)x + (y_0-k)y &= x_0^2 + y_0^2 -h x_0 - k y_0\
end{align}
answered May 16 '18 at 23:24
steven gregorysteven gregory
18k32258
$endgroup$
add a comment |
$begingroup$
Let the point of tangency be $P=(x_0, y_0)$. If the center of the circle is at $C=(h,k)$, then the line $overleftrightarrow{CP}$ is perpendicular to tangent line at the point $P$. The slope of the line $overleftrightarrow{CP}$ is $dfrac{y_0-k}{x_0-h}$, so the slope of the tangent line is $m=-dfrac{x_0-h}{y_0-k}$.
The equation of the tangent line is therefore
$$y - y_0 = -dfrac{x_0-h}{y_0-k}(x-x_0)$$
Just to satisfy my own curiosity, I want to write this equation is standard form.
begin{align}
y - y_0 &= -dfrac{x_0-h}{y_0-k}(x-x_0) \
(y_0-k)y - (y_0-k)y_0 &= -(x_0-h)x + (x_0-h)x_0 \
(x_0-h)x + (y_0-k)y &= (x_0-h)x_0 + (y_0-k)y_0 \
(x_0-h)x + (y_0-k)y &= x_0^2 + y_0^2 -h x_0 - k y_0\
end{align}
answered May 16 '18 at 23:24
steven gregorysteven gregory
18k32258
$endgroup$
add a comment |
$begingroup$
Let the point of tangency be $P=(x_0, y_0)$. If the center of the circle is at $C=(h,k)$, then the line $overleftrightarrow{CP}$ is perpendicular to tangent line at the point $P$. The slope of the line $overleftrightarrow{CP}$ is $dfrac{y_0-k}{x_0-h}$, so the slope of the tangent line is $m=-dfrac{x_0-h}{y_0-k}$.
The equation of the tangent line is therefore
$$y - y_0 = -dfrac{x_0-h}{y_0-k}(x-x_0)$$
Just to satisfy my own curiosity, I want to write this equation is standard form.
begin{align}
y - y_0 &= -dfrac{x_0-h}{y_0-k}(x-x_0) \
(y_0-k)y - (y_0-k)y_0 &= -(x_0-h)x + (x_0-h)x_0 \
(x_0-h)x + (y_0-k)y &= (x_0-h)x_0 + (y_0-k)y_0 \
(x_0-h)x + (y_0-k)y &= x_0^2 + y_0^2 -h x_0 - k y_0\
end{align}
answered May 16 '18 at 23:24
steven gregorysteven gregory
18k32258
$endgroup$
Let the point of tangency be $P=(x_0, y_0)$. If the center of the circle is at $C=(h,k)$, then the line $overleftrightarrow{CP}$ is perpendicular to tangent line at the point $P$. The slope of the line $overleftrightarrow{CP}$ is $dfrac{y_0-k}{x_0-h}$, so the slope of the tangent line is $m=-dfrac{x_0-h}{y_0-k}$.
The equation of the tangent line is therefore
$$y - y_0 = -dfrac{x_0-h}{y_0-k}(x-x_0)$$
Just to satisfy my own curiosity, I want to write this equation is standard form.
begin{align}
y - y_0 &= -dfrac{x_0-h}{y_0-k}(x-x_0) \
(y_0-k)y - (y_0-k)y_0 &= -(x_0-h)x + (x_0-h)x_0 \
(x_0-h)x + (y_0-k)y &= (x_0-h)x_0 + (y_0-k)y_0 \
(x_0-h)x + (y_0-k)y &= x_0^2 + y_0^2 -h x_0 - k y_0\
end{align}
answered May 16 '18 at 23:24
steven gregorysteven gregory
18k32258
edited May 16 '18 at 23:36
answered May 16 '18 at 23:24
steven gregorysteven gregory
18k32258
answered May 16 '18 at 23:24
steven gregorysteven gregory
18k32258
answered May 16 '18 at 23:24
steven gregorysteven gregory
18k32258
18k32258
add a comment |
add a comment |
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