What is radius of smaller circle in this case? [closed]
There is one quarter circle have radius same as side of square. There is semi-circle that has diameter equal to side of square. Then find radius of circle which is tangent to these circle and side of square.
geometry
asked Nov 20 at 11:24
Angelus Mortis
1135
closed as off-topic by José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn Nov 20 at 15:59
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- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn
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There is one quarter circle have radius same as side of square. There is semi-circle that has diameter equal to side of square. Then find radius of circle which is tangent to these circle and side of square.
geometry
asked Nov 20 at 11:24
Angelus Mortis
1135
closed as off-topic by José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn Nov 20 at 15:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
There is one quarter circle have radius same as side of square. There is semi-circle that has diameter equal to side of square. Then find radius of circle which is tangent to these circle and side of square.
geometry
asked Nov 20 at 11:24
Angelus Mortis
1135
There is one quarter circle have radius same as side of square. There is semi-circle that has diameter equal to side of square. Then find radius of circle which is tangent to these circle and side of square.
geometry
geometry
asked Nov 20 at 11:24
Angelus Mortis
1135
asked Nov 20 at 11:24
Angelus Mortis
1135
asked Nov 20 at 11:24
Angelus Mortis
1135
asked Nov 20 at 11:24
Angelus Mortis
1135
asked Nov 20 at 11:24
Angelus Mortis
1135
1135
closed as off-topic by José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn Nov 20 at 15:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn Nov 20 at 15:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Christopher, supinf, Adrian Keister, ccorn
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
3 Answers
3
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oldest
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Hint: For each of the two large circles, construct the right triangle with hypotenuse connecting the center to the center of the small circle and legs parallel to the sides of the square.
Let $x$ denote the distance between the lower left corner of the square and the point of tangency between the bottom side and the small circle.
Pythagorus gives us two equations:
$$(4-r)^2+x^2=(4+r)^2quad & quad (2-r)^2+(4-x)^2=(2+r)^2$$
answered Nov 20 at 11:31
lulu
39k24677
This is an excellent hint, accessible even for student with a basic knowledge of (analytic) geometry (+1)
– user376343
Nov 20 at 13:37
add a comment |
A geometric construction based on the use of conic sections:
- Locus of the centers of circles touching two given circles is a hyperbola, whose foci are the centers of the given circles as explained here.
- Locus of the centers of circles touching a given circle and a straight line is a parabola. Its focus is located in the center of the given circle, and the directrix is parallel to the given line, the given line being equidistant to the focus and directrix.
In the enclosed figure is the focus $A$ and the straight line $CD.$ The directrix lies bellow, and is out of the picture.
GeoGebra gives a nice picture, but the computation of the radius based on this method would be rather boring.
answered Nov 20 at 14:00
user376343
2,7882822
add a comment |
Hint: My attempt would be to formulate the equations for the three circles and the line segement and then check if there is a common solution possible. (Two intersection points with the circles, one with the line)
Circle: $(x - x_0)^2 + (y - y_0)^2 = r^2$
Line: $A x + B y = C$
Looks close to Descartes theorem, by the way.
Update:
Choosing the lower left corner as origin we get the equations:
Large circle
$$
x^2 + (y - 4)^2 = 4^2 quadquad (L)
$$
Medium sized circle
$$
(x-4)^2 + (y-2)^2 = 2^2 quadquad (M)
$$
Small circle
$$
(x-x_0)^2 + (y-y_0)^2 = r^2 quadquad (S)
$$
Line segment
$$
x in [0, 4] quad y = 0 quadquad (LS)
$$
The small circle touches the line segment from above, if the center is
$$
(x_0, y_0) = (x_0, r)
$$
for some $x_0$.
answered Nov 20 at 11:32
mvw
31.3k22252
I thought about Descartes theorem too but it won't apply here. Anyways I quite didn't understand your hint. sorry
– Angelus Mortis
Nov 20 at 12:45
I will expand it a bit later today.
– mvw
Nov 20 at 15:40
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint: For each of the two large circles, construct the right triangle with hypotenuse connecting the center to the center of the small circle and legs parallel to the sides of the square.
Let $x$ denote the distance between the lower left corner of the square and the point of tangency between the bottom side and the small circle.
Pythagorus gives us two equations:
$$(4-r)^2+x^2=(4+r)^2quad & quad (2-r)^2+(4-x)^2=(2+r)^2$$
answered Nov 20 at 11:31
lulu
39k24677
This is an excellent hint, accessible even for student with a basic knowledge of (analytic) geometry (+1)
– user376343
Nov 20 at 13:37
add a comment |
Hint: For each of the two large circles, construct the right triangle with hypotenuse connecting the center to the center of the small circle and legs parallel to the sides of the square.
Let $x$ denote the distance between the lower left corner of the square and the point of tangency between the bottom side and the small circle.
Pythagorus gives us two equations:
$$(4-r)^2+x^2=(4+r)^2quad & quad (2-r)^2+(4-x)^2=(2+r)^2$$
answered Nov 20 at 11:31
lulu
39k24677
This is an excellent hint, accessible even for student with a basic knowledge of (analytic) geometry (+1)
– user376343
Nov 20 at 13:37
add a comment |
Hint: For each of the two large circles, construct the right triangle with hypotenuse connecting the center to the center of the small circle and legs parallel to the sides of the square.
Let $x$ denote the distance between the lower left corner of the square and the point of tangency between the bottom side and the small circle.
Pythagorus gives us two equations:
$$(4-r)^2+x^2=(4+r)^2quad & quad (2-r)^2+(4-x)^2=(2+r)^2$$
answered Nov 20 at 11:31
lulu
39k24677
Hint: For each of the two large circles, construct the right triangle with hypotenuse connecting the center to the center of the small circle and legs parallel to the sides of the square.
Let $x$ denote the distance between the lower left corner of the square and the point of tangency between the bottom side and the small circle.
Pythagorus gives us two equations:
$$(4-r)^2+x^2=(4+r)^2quad & quad (2-r)^2+(4-x)^2=(2+r)^2$$
answered Nov 20 at 11:31
lulu
39k24677
edited Nov 20 at 13:50
answered Nov 20 at 11:31
lulu
39k24677
answered Nov 20 at 11:31
lulu
39k24677
answered Nov 20 at 11:31
lulu
39k24677
39k24677
This is an excellent hint, accessible even for student with a basic knowledge of (analytic) geometry (+1)
– user376343
Nov 20 at 13:37
add a comment |
This is an excellent hint, accessible even for student with a basic knowledge of (analytic) geometry (+1)
– user376343
Nov 20 at 13:37
This is an excellent hint, accessible even for student with a basic knowledge of (analytic) geometry (+1)
– user376343
Nov 20 at 13:37
This is an excellent hint, accessible even for student with a basic knowledge of (analytic) geometry (+1)
– user376343
Nov 20 at 13:37
add a comment |
A geometric construction based on the use of conic sections:
- Locus of the centers of circles touching two given circles is a hyperbola, whose foci are the centers of the given circles as explained here.
- Locus of the centers of circles touching a given circle and a straight line is a parabola. Its focus is located in the center of the given circle, and the directrix is parallel to the given line, the given line being equidistant to the focus and directrix.
In the enclosed figure is the focus $A$ and the straight line $CD.$ The directrix lies bellow, and is out of the picture.
GeoGebra gives a nice picture, but the computation of the radius based on this method would be rather boring.
answered Nov 20 at 14:00
user376343
2,7882822
add a comment |
A geometric construction based on the use of conic sections:
- Locus of the centers of circles touching two given circles is a hyperbola, whose foci are the centers of the given circles as explained here.
- Locus of the centers of circles touching a given circle and a straight line is a parabola. Its focus is located in the center of the given circle, and the directrix is parallel to the given line, the given line being equidistant to the focus and directrix.
In the enclosed figure is the focus $A$ and the straight line $CD.$ The directrix lies bellow, and is out of the picture.
GeoGebra gives a nice picture, but the computation of the radius based on this method would be rather boring.
answered Nov 20 at 14:00
user376343
2,7882822
add a comment |
A geometric construction based on the use of conic sections:
- Locus of the centers of circles touching two given circles is a hyperbola, whose foci are the centers of the given circles as explained here.
- Locus of the centers of circles touching a given circle and a straight line is a parabola. Its focus is located in the center of the given circle, and the directrix is parallel to the given line, the given line being equidistant to the focus and directrix.
In the enclosed figure is the focus $A$ and the straight line $CD.$ The directrix lies bellow, and is out of the picture.
GeoGebra gives a nice picture, but the computation of the radius based on this method would be rather boring.
answered Nov 20 at 14:00
user376343
2,7882822
A geometric construction based on the use of conic sections:
- Locus of the centers of circles touching two given circles is a hyperbola, whose foci are the centers of the given circles as explained here.
- Locus of the centers of circles touching a given circle and a straight line is a parabola. Its focus is located in the center of the given circle, and the directrix is parallel to the given line, the given line being equidistant to the focus and directrix.
In the enclosed figure is the focus $A$ and the straight line $CD.$ The directrix lies bellow, and is out of the picture.
GeoGebra gives a nice picture, but the computation of the radius based on this method would be rather boring.
answered Nov 20 at 14:00
user376343
2,7882822
answered Nov 20 at 14:00
user376343
2,7882822
answered Nov 20 at 14:00
user376343
2,7882822
answered Nov 20 at 14:00
user376343
2,7882822
2,7882822
add a comment |
add a comment |
Hint: My attempt would be to formulate the equations for the three circles and the line segement and then check if there is a common solution possible. (Two intersection points with the circles, one with the line)
Circle: $(x - x_0)^2 + (y - y_0)^2 = r^2$
Line: $A x + B y = C$
Looks close to Descartes theorem, by the way.
Update:
Choosing the lower left corner as origin we get the equations:
Large circle
$$
x^2 + (y - 4)^2 = 4^2 quadquad (L)
$$
Medium sized circle
$$
(x-4)^2 + (y-2)^2 = 2^2 quadquad (M)
$$
Small circle
$$
(x-x_0)^2 + (y-y_0)^2 = r^2 quadquad (S)
$$
Line segment
$$
x in [0, 4] quad y = 0 quadquad (LS)
$$
The small circle touches the line segment from above, if the center is
$$
(x_0, y_0) = (x_0, r)
$$
for some $x_0$.
answered Nov 20 at 11:32
mvw
31.3k22252
I thought about Descartes theorem too but it won't apply here. Anyways I quite didn't understand your hint. sorry
– Angelus Mortis
Nov 20 at 12:45
I will expand it a bit later today.
– mvw
Nov 20 at 15:40
add a comment |
Hint: My attempt would be to formulate the equations for the three circles and the line segement and then check if there is a common solution possible. (Two intersection points with the circles, one with the line)
Circle: $(x - x_0)^2 + (y - y_0)^2 = r^2$
Line: $A x + B y = C$
Looks close to Descartes theorem, by the way.
Update:
Choosing the lower left corner as origin we get the equations:
Large circle
$$
x^2 + (y - 4)^2 = 4^2 quadquad (L)
$$
Medium sized circle
$$
(x-4)^2 + (y-2)^2 = 2^2 quadquad (M)
$$
Small circle
$$
(x-x_0)^2 + (y-y_0)^2 = r^2 quadquad (S)
$$
Line segment
$$
x in [0, 4] quad y = 0 quadquad (LS)
$$
The small circle touches the line segment from above, if the center is
$$
(x_0, y_0) = (x_0, r)
$$
for some $x_0$.
answered Nov 20 at 11:32
mvw
31.3k22252
I thought about Descartes theorem too but it won't apply here. Anyways I quite didn't understand your hint. sorry
– Angelus Mortis
Nov 20 at 12:45
I will expand it a bit later today.
– mvw
Nov 20 at 15:40
add a comment |
Hint: My attempt would be to formulate the equations for the three circles and the line segement and then check if there is a common solution possible. (Two intersection points with the circles, one with the line)
Circle: $(x - x_0)^2 + (y - y_0)^2 = r^2$
Line: $A x + B y = C$
Looks close to Descartes theorem, by the way.
Update:
Choosing the lower left corner as origin we get the equations:
Large circle
$$
x^2 + (y - 4)^2 = 4^2 quadquad (L)
$$
Medium sized circle
$$
(x-4)^2 + (y-2)^2 = 2^2 quadquad (M)
$$
Small circle
$$
(x-x_0)^2 + (y-y_0)^2 = r^2 quadquad (S)
$$
Line segment
$$
x in [0, 4] quad y = 0 quadquad (LS)
$$
The small circle touches the line segment from above, if the center is
$$
(x_0, y_0) = (x_0, r)
$$
for some $x_0$.
answered Nov 20 at 11:32
mvw
31.3k22252
Hint: My attempt would be to formulate the equations for the three circles and the line segement and then check if there is a common solution possible. (Two intersection points with the circles, one with the line)
Circle: $(x - x_0)^2 + (y - y_0)^2 = r^2$
Line: $A x + B y = C$
Looks close to Descartes theorem, by the way.
Update:
Choosing the lower left corner as origin we get the equations:
Large circle
$$
x^2 + (y - 4)^2 = 4^2 quadquad (L)
$$
Medium sized circle
$$
(x-4)^2 + (y-2)^2 = 2^2 quadquad (M)
$$
Small circle
$$
(x-x_0)^2 + (y-y_0)^2 = r^2 quadquad (S)
$$
Line segment
$$
x in [0, 4] quad y = 0 quadquad (LS)
$$
The small circle touches the line segment from above, if the center is
$$
(x_0, y_0) = (x_0, r)
$$
for some $x_0$.
answered Nov 20 at 11:32
mvw
31.3k22252
edited Nov 20 at 15:39
answered Nov 20 at 11:32
mvw
31.3k22252
answered Nov 20 at 11:32
mvw
31.3k22252
answered Nov 20 at 11:32
mvw
31.3k22252
31.3k22252
I thought about Descartes theorem too but it won't apply here. Anyways I quite didn't understand your hint. sorry
– Angelus Mortis
Nov 20 at 12:45
I will expand it a bit later today.
– mvw
Nov 20 at 15:40
add a comment |
I thought about Descartes theorem too but it won't apply here. Anyways I quite didn't understand your hint. sorry
– Angelus Mortis
Nov 20 at 12:45
I will expand it a bit later today.
– mvw
Nov 20 at 15:40
I thought about Descartes theorem too but it won't apply here. Anyways I quite didn't understand your hint. sorry
– Angelus Mortis
Nov 20 at 12:45
I thought about Descartes theorem too but it won't apply here. Anyways I quite didn't understand your hint. sorry
– Angelus Mortis
Nov 20 at 12:45
I will expand it a bit later today.
– mvw
Nov 20 at 15:40
I will expand it a bit later today.
– mvw
Nov 20 at 15:40
add a comment |
