How to find a common tangent line given two circles?
The equations of the two circles are $x^2+y^2=36$ and $(x-5)^2 +y^2=16$. The problem asks to find a common tangent line in point-slope form.
I've tried drawing a diagram and finding the distance between the points of tangency, but that did not help in finding a point of tangency or the slope of the lines.
geometry algebraic-geometry
edited Mar 21 '17 at 23:53
zoli
16.4k41643
asked Mar 21 '17 at 23:42
lumpiestspoon3
111
add a comment |
The equations of the two circles are $x^2+y^2=36$ and $(x-5)^2 +y^2=16$. The problem asks to find a common tangent line in point-slope form.
I've tried drawing a diagram and finding the distance between the points of tangency, but that did not help in finding a point of tangency or the slope of the lines.
geometry algebraic-geometry
edited Mar 21 '17 at 23:53
zoli
16.4k41643
asked Mar 21 '17 at 23:42
lumpiestspoon3
111
Draw the common tangent & think about where it intersects the $x$ axis. Now draw the normals (from the centers of the circles) to their respective tangent points. You should have two similar right angled triangles ? now you can set up a ratio ? ... more hints on request
– Donald Splutterwit
Mar 22 '17 at 0:15
add a comment |
The equations of the two circles are $x^2+y^2=36$ and $(x-5)^2 +y^2=16$. The problem asks to find a common tangent line in point-slope form.
I've tried drawing a diagram and finding the distance between the points of tangency, but that did not help in finding a point of tangency or the slope of the lines.
geometry algebraic-geometry
edited Mar 21 '17 at 23:53
zoli
16.4k41643
asked Mar 21 '17 at 23:42
lumpiestspoon3
111
The equations of the two circles are $x^2+y^2=36$ and $(x-5)^2 +y^2=16$. The problem asks to find a common tangent line in point-slope form.
I've tried drawing a diagram and finding the distance between the points of tangency, but that did not help in finding a point of tangency or the slope of the lines.
geometry algebraic-geometry
geometry algebraic-geometry
edited Mar 21 '17 at 23:53
zoli
16.4k41643
asked Mar 21 '17 at 23:42
lumpiestspoon3
111
edited Mar 21 '17 at 23:53
zoli
16.4k41643
asked Mar 21 '17 at 23:42
lumpiestspoon3
111
edited Mar 21 '17 at 23:53
zoli
16.4k41643
edited Mar 21 '17 at 23:53
zoli
16.4k41643
edited Mar 21 '17 at 23:53
zoli
16.4k41643
16.4k41643
asked Mar 21 '17 at 23:42
lumpiestspoon3
111
asked Mar 21 '17 at 23:42
lumpiestspoon3
111
asked Mar 21 '17 at 23:42
lumpiestspoon3
111
111
Draw the common tangent & think about where it intersects the $x$ axis. Now draw the normals (from the centers of the circles) to their respective tangent points. You should have two similar right angled triangles ? now you can set up a ratio ? ... more hints on request
– Donald Splutterwit
Mar 22 '17 at 0:15
add a comment |
Draw the common tangent & think about where it intersects the $x$ axis. Now draw the normals (from the centers of the circles) to their respective tangent points. You should have two similar right angled triangles ? now you can set up a ratio ? ... more hints on request
– Donald Splutterwit
Mar 22 '17 at 0:15
Draw the common tangent & think about where it intersects the $x$ axis. Now draw the normals (from the centers of the circles) to their respective tangent points. You should have two similar right angled triangles ? now you can set up a ratio ? ... more hints on request
– Donald Splutterwit
Mar 22 '17 at 0:15
Draw the common tangent & think about where it intersects the $x$ axis. Now draw the normals (from the centers of the circles) to their respective tangent points. You should have two similar right angled triangles ? now you can set up a ratio ? ... more hints on request
– Donald Splutterwit
Mar 22 '17 at 0:15
add a comment |
2 Answers
2
active
oldest
votes
A hint (after Donald Splutterwit):
(Sorry... one cannot post graphics in a comment...)
So solve for $x$:
${4 over x} = {6 over 5+x}$
to find $x = 10$.
Now what?....
answered Mar 22 '17 at 0:22
David G. Stork
9,56721232
add a comment |
The two tangents either side of the circles meet at a point, and the line through the circle centres bisects that meeting angle. The distance to that point is in ratio to the circle diameters, so given that the circle centres given here are $5$ units apart, and are radius $6$ and $4$, the meeting point of the two tangents will be $10$ units beyond the $4$ radius circle, giving a point on both tangents of $(15,0)$.
Then the slope can be calculated from a right triangle to the touch point on the near triangle as $pmfrac 4a $ given $a^2 = 10^2-4^2$.
answered Mar 22 '17 at 0:30
Joffan
32.2k43169
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
oldest
votes
A hint (after Donald Splutterwit):
(Sorry... one cannot post graphics in a comment...)
So solve for $x$:
${4 over x} = {6 over 5+x}$
to find $x = 10$.
Now what?....
answered Mar 22 '17 at 0:22
David G. Stork
9,56721232
add a comment |
A hint (after Donald Splutterwit):
(Sorry... one cannot post graphics in a comment...)
So solve for $x$:
${4 over x} = {6 over 5+x}$
to find $x = 10$.
Now what?....
answered Mar 22 '17 at 0:22
David G. Stork
9,56721232
add a comment |
A hint (after Donald Splutterwit):
(Sorry... one cannot post graphics in a comment...)
So solve for $x$:
${4 over x} = {6 over 5+x}$
to find $x = 10$.
Now what?....
answered Mar 22 '17 at 0:22
David G. Stork
9,56721232
A hint (after Donald Splutterwit):
(Sorry... one cannot post graphics in a comment...)
So solve for $x$:
${4 over x} = {6 over 5+x}$
to find $x = 10$.
Now what?....
answered Mar 22 '17 at 0:22
David G. Stork
9,56721232
answered Mar 22 '17 at 0:22
David G. Stork
9,56721232
answered Mar 22 '17 at 0:22
David G. Stork
9,56721232
answered Mar 22 '17 at 0:22
David G. Stork
9,56721232
9,56721232
add a comment |
add a comment |
The two tangents either side of the circles meet at a point, and the line through the circle centres bisects that meeting angle. The distance to that point is in ratio to the circle diameters, so given that the circle centres given here are $5$ units apart, and are radius $6$ and $4$, the meeting point of the two tangents will be $10$ units beyond the $4$ radius circle, giving a point on both tangents of $(15,0)$.
Then the slope can be calculated from a right triangle to the touch point on the near triangle as $pmfrac 4a $ given $a^2 = 10^2-4^2$.
answered Mar 22 '17 at 0:30
Joffan
32.2k43169
add a comment |
The two tangents either side of the circles meet at a point, and the line through the circle centres bisects that meeting angle. The distance to that point is in ratio to the circle diameters, so given that the circle centres given here are $5$ units apart, and are radius $6$ and $4$, the meeting point of the two tangents will be $10$ units beyond the $4$ radius circle, giving a point on both tangents of $(15,0)$.
Then the slope can be calculated from a right triangle to the touch point on the near triangle as $pmfrac 4a $ given $a^2 = 10^2-4^2$.
answered Mar 22 '17 at 0:30
Joffan
32.2k43169
add a comment |
The two tangents either side of the circles meet at a point, and the line through the circle centres bisects that meeting angle. The distance to that point is in ratio to the circle diameters, so given that the circle centres given here are $5$ units apart, and are radius $6$ and $4$, the meeting point of the two tangents will be $10$ units beyond the $4$ radius circle, giving a point on both tangents of $(15,0)$.
Then the slope can be calculated from a right triangle to the touch point on the near triangle as $pmfrac 4a $ given $a^2 = 10^2-4^2$.
answered Mar 22 '17 at 0:30
Joffan
32.2k43169
The two tangents either side of the circles meet at a point, and the line through the circle centres bisects that meeting angle. The distance to that point is in ratio to the circle diameters, so given that the circle centres given here are $5$ units apart, and are radius $6$ and $4$, the meeting point of the two tangents will be $10$ units beyond the $4$ radius circle, giving a point on both tangents of $(15,0)$.
Then the slope can be calculated from a right triangle to the touch point on the near triangle as $pmfrac 4a $ given $a^2 = 10^2-4^2$.
answered Mar 22 '17 at 0:30
Joffan
32.2k43169
answered Mar 22 '17 at 0:30
Joffan
32.2k43169
answered Mar 22 '17 at 0:30
Joffan
32.2k43169
answered Mar 22 '17 at 0:30
Joffan
32.2k43169
32.2k43169
add a comment |
add a comment |
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Draw the common tangent & think about where it intersects the $x$ axis. Now draw the normals (from the centers of the circles) to their respective tangent points. You should have two similar right angled triangles ? now you can set up a ratio ? ... more hints on request
– Donald Splutterwit
Mar 22 '17 at 0:15