Can we prove that two circles lie on a sphere?
$begingroup$
There is a problem in my book
In the question the circles are given in the form
$S+kP=0$
where $S$ is plane and $P$ is a plane and $k$ is real number.
The question asks us to prove that the circles lie on the same sphere and to find the equation of the sphere .
I can answer only the second part.
I asked my teacher and he said that we can't prove the first part.
Any thoughts or suggestions??
Thanks.
surfaces spheres
edited Jan 7 at 18:10
Robert Z
101k1070143
asked Jan 7 at 17:44
PN DasPN Das
157
$endgroup$
add a comment |
$begingroup$
There is a problem in my book
In the question the circles are given in the form
$S+kP=0$
where $S$ is plane and $P$ is a plane and $k$ is real number.
The question asks us to prove that the circles lie on the same sphere and to find the equation of the sphere .
I can answer only the second part.
I asked my teacher and he said that we can't prove the first part.
Any thoughts or suggestions??
Thanks.
surfaces spheres
edited Jan 7 at 18:10
Robert Z
101k1070143
asked Jan 7 at 17:44
PN DasPN Das
157
$endgroup$
$begingroup$
It might help to give the title and author of your textbook. By doing only the second part you seem to mean finding the equation of the sphere. If you have done that, doesn't it become fairly straightforward to check that the circles lie on the sphere by comparing the coordinates?
$endgroup$
– hardmath
Jan 9 at 3:17
add a comment |
$begingroup$
There is a problem in my book
In the question the circles are given in the form
$S+kP=0$
where $S$ is plane and $P$ is a plane and $k$ is real number.
The question asks us to prove that the circles lie on the same sphere and to find the equation of the sphere .
I can answer only the second part.
I asked my teacher and he said that we can't prove the first part.
Any thoughts or suggestions??
Thanks.
surfaces spheres
edited Jan 7 at 18:10
Robert Z
101k1070143
asked Jan 7 at 17:44
PN DasPN Das
157
$endgroup$
There is a problem in my book
In the question the circles are given in the form
$S+kP=0$
where $S$ is plane and $P$ is a plane and $k$ is real number.
The question asks us to prove that the circles lie on the same sphere and to find the equation of the sphere .
I can answer only the second part.
I asked my teacher and he said that we can't prove the first part.
Any thoughts or suggestions??
Thanks.
surfaces spheres
surfaces spheres
edited Jan 7 at 18:10
Robert Z
101k1070143
asked Jan 7 at 17:44
PN DasPN Das
157
edited Jan 7 at 18:10
Robert Z
101k1070143
asked Jan 7 at 17:44
PN DasPN Das
157
edited Jan 7 at 18:10
Robert Z
101k1070143
edited Jan 7 at 18:10
Robert Z
101k1070143
edited Jan 7 at 18:10
Robert Z
101k1070143
101k1070143
asked Jan 7 at 17:44
PN DasPN Das
157
asked Jan 7 at 17:44
PN DasPN Das
157
asked Jan 7 at 17:44
PN DasPN Das
157
157
$begingroup$
It might help to give the title and author of your textbook. By doing only the second part you seem to mean finding the equation of the sphere. If you have done that, doesn't it become fairly straightforward to check that the circles lie on the sphere by comparing the coordinates?
$endgroup$
– hardmath
Jan 9 at 3:17
add a comment |
$begingroup$
It might help to give the title and author of your textbook. By doing only the second part you seem to mean finding the equation of the sphere. If you have done that, doesn't it become fairly straightforward to check that the circles lie on the sphere by comparing the coordinates?
$endgroup$
– hardmath
Jan 9 at 3:17
$begingroup$
It might help to give the title and author of your textbook. By doing only the second part you seem to mean finding the equation of the sphere. If you have done that, doesn't it become fairly straightforward to check that the circles lie on the sphere by comparing the coordinates?
$endgroup$
– hardmath
Jan 9 at 3:17
$begingroup$
It might help to give the title and author of your textbook. By doing only the second part you seem to mean finding the equation of the sphere. If you have done that, doesn't it become fairly straightforward to check that the circles lie on the sphere by comparing the coordinates?
$endgroup$
– hardmath
Jan 9 at 3:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us type all data here.
$$
begin{aligned}
S_1(x,y,z) &= x^2 +y^2+z^2 - 2x+3y+4z-5 ,\
P_1(x,y,z) &= 5y+6z+1 ,\[2mm]
S_2(x,y,z) &= x^2 +y^2+z^2 - 3x-4y+5z-6 ,\
P_2(x,y,z) &= x+2y-7z ,\[2mm]
(C_1)&={ (x,y,z) : S_1(x,y,z)=P_1(x,y,z)=0 } ,\
(C_2)&={ (x,y,z) : S_2(x,y,z)=P_2(x,y,z)=0 } .
end{aligned}
$$
The following solution supports the geometric intuition.
$S_1=0$ describes a sphere centered in $Omega_1=displaystyle left(1,-frac 32,-2right)$ with radius $frac 72$. We cut it with a plane, $P_1=0$, getting a circle.
If this circle is on some other sphere, then this other sphere is centered in a point $Omega$ on the normal form $Omega_1$ to the plane. We already see the plane and its normal, so $Omega$ is on the line with points parametrized as $displaystyle left(1,-frac 32,-2right)+lambda(0,5,6)$, $lambdainBbb R$.
The same game for the other circle.
$S_2=0$ describes a sphere centered in $Omega_2=displaystyle left(frac 32,2, -frac52right)$ with radius $4$. We cut it with a second plane, $P_2=0$, getting a circle.
If this circle is on the same sphere above, then $Omega$ is on the normal form $Omega_2$ to the plane. So $Omega$ is on the line with points parametrized as $displaystyle left(frac 32,2,-frac52right)+mu(1,2,-7)$, $muinBbb R$.
Do these two lines intersect? Yes, $lambda=frac 12$, $mu=-frac 12$, lead to $Omega=(1,1,1)$. So far we land so far as using the hint. ($lambda,mu$ are up to a factor two some $lambda_1,lambda_2$ special values from the hint.)
We still have to check the "radius match". Here we should go back to algebraic geometry.
A point is on $(C_1)$ if it is satisfying for all $lambda_1$ the equation $S_1+lambda_1 P_1=0$.
A point is on $(C_2)$ if it is satisfying for all $lambda_2$ the equation $S_2+lambda_2 P_2=0$.
Now use $lambda_1=2lambda$, $lambda_2=2mu$ (some "equation dedoubling" is the reason for short) and check that we have the same equation. (This corresponds to a "radius match", algebraically to a match of the constant coefficient.)
answered Jan 7 at 18:56
dan_fuleadan_fulea
6,9681413
$endgroup$
$begingroup$
Thank you very much sir ! I didn't know this geometrical intuition . But what if the equations of circles are given in their general form like x^2+y^2=r^2 . How can we prove that they lie on same sphere.
$endgroup$
– PN Das
Jan 8 at 15:34
$begingroup$
@PN DAS: The two circles in the problem were given as intersection of a sphere with a plane. For instance, the circle $(C_1)$ is the intersection of the sphere with equation $S_1=0$, with the plane $P_1=0$. Here,$$begin{aligned}S_1(x,y,z) &= x^2 +y^2+z^2 - 2x + 3y + 4z - 5\&=(x-1)^2+(y+3/2)^2+(z+2)^2-49/4end{aligned}$$and from it we extract the center and the radius. A circle cannot be given in 3D by only one equation. An equation like $(x-a)^2+(y-b)^2=r^2$ in 2D, i.e. $(x,y)inBbb R^2$, is a circle, centered in $(a,b)$, radius $r$. In 3D we need one more term² & it's a sphere as above.
$endgroup$
– dan_fulea
Jan 8 at 17:46
$begingroup$
An equation like $x^2+y^2=r^2$ (or more general $(x-a)^2+(y-b)^2=r^2$), seen in three dimensions, i.e. considering the set of all $(x,y,z)$ that satisfy the equation, is a cylinder.
$endgroup$
– dan_fulea
Jan 8 at 17:49
add a comment |
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$begingroup$
Let us type all data here.
$$
begin{aligned}
S_1(x,y,z) &= x^2 +y^2+z^2 - 2x+3y+4z-5 ,\
P_1(x,y,z) &= 5y+6z+1 ,\[2mm]
S_2(x,y,z) &= x^2 +y^2+z^2 - 3x-4y+5z-6 ,\
P_2(x,y,z) &= x+2y-7z ,\[2mm]
(C_1)&={ (x,y,z) : S_1(x,y,z)=P_1(x,y,z)=0 } ,\
(C_2)&={ (x,y,z) : S_2(x,y,z)=P_2(x,y,z)=0 } .
end{aligned}
$$
The following solution supports the geometric intuition.
$S_1=0$ describes a sphere centered in $Omega_1=displaystyle left(1,-frac 32,-2right)$ with radius $frac 72$. We cut it with a plane, $P_1=0$, getting a circle.
If this circle is on some other sphere, then this other sphere is centered in a point $Omega$ on the normal form $Omega_1$ to the plane. We already see the plane and its normal, so $Omega$ is on the line with points parametrized as $displaystyle left(1,-frac 32,-2right)+lambda(0,5,6)$, $lambdainBbb R$.
The same game for the other circle.
$S_2=0$ describes a sphere centered in $Omega_2=displaystyle left(frac 32,2, -frac52right)$ with radius $4$. We cut it with a second plane, $P_2=0$, getting a circle.
If this circle is on the same sphere above, then $Omega$ is on the normal form $Omega_2$ to the plane. So $Omega$ is on the line with points parametrized as $displaystyle left(frac 32,2,-frac52right)+mu(1,2,-7)$, $muinBbb R$.
Do these two lines intersect? Yes, $lambda=frac 12$, $mu=-frac 12$, lead to $Omega=(1,1,1)$. So far we land so far as using the hint. ($lambda,mu$ are up to a factor two some $lambda_1,lambda_2$ special values from the hint.)
We still have to check the "radius match". Here we should go back to algebraic geometry.
A point is on $(C_1)$ if it is satisfying for all $lambda_1$ the equation $S_1+lambda_1 P_1=0$.
A point is on $(C_2)$ if it is satisfying for all $lambda_2$ the equation $S_2+lambda_2 P_2=0$.
Now use $lambda_1=2lambda$, $lambda_2=2mu$ (some "equation dedoubling" is the reason for short) and check that we have the same equation. (This corresponds to a "radius match", algebraically to a match of the constant coefficient.)
answered Jan 7 at 18:56
dan_fuleadan_fulea
6,9681413
$endgroup$
$begingroup$
Thank you very much sir ! I didn't know this geometrical intuition . But what if the equations of circles are given in their general form like x^2+y^2=r^2 . How can we prove that they lie on same sphere.
$endgroup$
– PN Das
Jan 8 at 15:34
$begingroup$
@PN DAS: The two circles in the problem were given as intersection of a sphere with a plane. For instance, the circle $(C_1)$ is the intersection of the sphere with equation $S_1=0$, with the plane $P_1=0$. Here,$$begin{aligned}S_1(x,y,z) &= x^2 +y^2+z^2 - 2x + 3y + 4z - 5\&=(x-1)^2+(y+3/2)^2+(z+2)^2-49/4end{aligned}$$and from it we extract the center and the radius. A circle cannot be given in 3D by only one equation. An equation like $(x-a)^2+(y-b)^2=r^2$ in 2D, i.e. $(x,y)inBbb R^2$, is a circle, centered in $(a,b)$, radius $r$. In 3D we need one more term² & it's a sphere as above.
$endgroup$
– dan_fulea
Jan 8 at 17:46
$begingroup$
An equation like $x^2+y^2=r^2$ (or more general $(x-a)^2+(y-b)^2=r^2$), seen in three dimensions, i.e. considering the set of all $(x,y,z)$ that satisfy the equation, is a cylinder.
$endgroup$
– dan_fulea
Jan 8 at 17:49
add a comment |
$begingroup$
Let us type all data here.
$$
begin{aligned}
S_1(x,y,z) &= x^2 +y^2+z^2 - 2x+3y+4z-5 ,\
P_1(x,y,z) &= 5y+6z+1 ,\[2mm]
S_2(x,y,z) &= x^2 +y^2+z^2 - 3x-4y+5z-6 ,\
P_2(x,y,z) &= x+2y-7z ,\[2mm]
(C_1)&={ (x,y,z) : S_1(x,y,z)=P_1(x,y,z)=0 } ,\
(C_2)&={ (x,y,z) : S_2(x,y,z)=P_2(x,y,z)=0 } .
end{aligned}
$$
The following solution supports the geometric intuition.
$S_1=0$ describes a sphere centered in $Omega_1=displaystyle left(1,-frac 32,-2right)$ with radius $frac 72$. We cut it with a plane, $P_1=0$, getting a circle.
If this circle is on some other sphere, then this other sphere is centered in a point $Omega$ on the normal form $Omega_1$ to the plane. We already see the plane and its normal, so $Omega$ is on the line with points parametrized as $displaystyle left(1,-frac 32,-2right)+lambda(0,5,6)$, $lambdainBbb R$.
The same game for the other circle.
$S_2=0$ describes a sphere centered in $Omega_2=displaystyle left(frac 32,2, -frac52right)$ with radius $4$. We cut it with a second plane, $P_2=0$, getting a circle.
If this circle is on the same sphere above, then $Omega$ is on the normal form $Omega_2$ to the plane. So $Omega$ is on the line with points parametrized as $displaystyle left(frac 32,2,-frac52right)+mu(1,2,-7)$, $muinBbb R$.
Do these two lines intersect? Yes, $lambda=frac 12$, $mu=-frac 12$, lead to $Omega=(1,1,1)$. So far we land so far as using the hint. ($lambda,mu$ are up to a factor two some $lambda_1,lambda_2$ special values from the hint.)
We still have to check the "radius match". Here we should go back to algebraic geometry.
A point is on $(C_1)$ if it is satisfying for all $lambda_1$ the equation $S_1+lambda_1 P_1=0$.
A point is on $(C_2)$ if it is satisfying for all $lambda_2$ the equation $S_2+lambda_2 P_2=0$.
Now use $lambda_1=2lambda$, $lambda_2=2mu$ (some "equation dedoubling" is the reason for short) and check that we have the same equation. (This corresponds to a "radius match", algebraically to a match of the constant coefficient.)
answered Jan 7 at 18:56
dan_fuleadan_fulea
6,9681413
$endgroup$
$begingroup$
Thank you very much sir ! I didn't know this geometrical intuition . But what if the equations of circles are given in their general form like x^2+y^2=r^2 . How can we prove that they lie on same sphere.
$endgroup$
– PN Das
Jan 8 at 15:34
$begingroup$
@PN DAS: The two circles in the problem were given as intersection of a sphere with a plane. For instance, the circle $(C_1)$ is the intersection of the sphere with equation $S_1=0$, with the plane $P_1=0$. Here,$$begin{aligned}S_1(x,y,z) &= x^2 +y^2+z^2 - 2x + 3y + 4z - 5\&=(x-1)^2+(y+3/2)^2+(z+2)^2-49/4end{aligned}$$and from it we extract the center and the radius. A circle cannot be given in 3D by only one equation. An equation like $(x-a)^2+(y-b)^2=r^2$ in 2D, i.e. $(x,y)inBbb R^2$, is a circle, centered in $(a,b)$, radius $r$. In 3D we need one more term² & it's a sphere as above.
$endgroup$
– dan_fulea
Jan 8 at 17:46
$begingroup$
An equation like $x^2+y^2=r^2$ (or more general $(x-a)^2+(y-b)^2=r^2$), seen in three dimensions, i.e. considering the set of all $(x,y,z)$ that satisfy the equation, is a cylinder.
$endgroup$
– dan_fulea
Jan 8 at 17:49
add a comment |
$begingroup$
Let us type all data here.
$$
begin{aligned}
S_1(x,y,z) &= x^2 +y^2+z^2 - 2x+3y+4z-5 ,\
P_1(x,y,z) &= 5y+6z+1 ,\[2mm]
S_2(x,y,z) &= x^2 +y^2+z^2 - 3x-4y+5z-6 ,\
P_2(x,y,z) &= x+2y-7z ,\[2mm]
(C_1)&={ (x,y,z) : S_1(x,y,z)=P_1(x,y,z)=0 } ,\
(C_2)&={ (x,y,z) : S_2(x,y,z)=P_2(x,y,z)=0 } .
end{aligned}
$$
The following solution supports the geometric intuition.
$S_1=0$ describes a sphere centered in $Omega_1=displaystyle left(1,-frac 32,-2right)$ with radius $frac 72$. We cut it with a plane, $P_1=0$, getting a circle.
If this circle is on some other sphere, then this other sphere is centered in a point $Omega$ on the normal form $Omega_1$ to the plane. We already see the plane and its normal, so $Omega$ is on the line with points parametrized as $displaystyle left(1,-frac 32,-2right)+lambda(0,5,6)$, $lambdainBbb R$.
The same game for the other circle.
$S_2=0$ describes a sphere centered in $Omega_2=displaystyle left(frac 32,2, -frac52right)$ with radius $4$. We cut it with a second plane, $P_2=0$, getting a circle.
If this circle is on the same sphere above, then $Omega$ is on the normal form $Omega_2$ to the plane. So $Omega$ is on the line with points parametrized as $displaystyle left(frac 32,2,-frac52right)+mu(1,2,-7)$, $muinBbb R$.
Do these two lines intersect? Yes, $lambda=frac 12$, $mu=-frac 12$, lead to $Omega=(1,1,1)$. So far we land so far as using the hint. ($lambda,mu$ are up to a factor two some $lambda_1,lambda_2$ special values from the hint.)
We still have to check the "radius match". Here we should go back to algebraic geometry.
A point is on $(C_1)$ if it is satisfying for all $lambda_1$ the equation $S_1+lambda_1 P_1=0$.
A point is on $(C_2)$ if it is satisfying for all $lambda_2$ the equation $S_2+lambda_2 P_2=0$.
Now use $lambda_1=2lambda$, $lambda_2=2mu$ (some "equation dedoubling" is the reason for short) and check that we have the same equation. (This corresponds to a "radius match", algebraically to a match of the constant coefficient.)
answered Jan 7 at 18:56
dan_fuleadan_fulea
6,9681413
$endgroup$
Let us type all data here.
$$
begin{aligned}
S_1(x,y,z) &= x^2 +y^2+z^2 - 2x+3y+4z-5 ,\
P_1(x,y,z) &= 5y+6z+1 ,\[2mm]
S_2(x,y,z) &= x^2 +y^2+z^2 - 3x-4y+5z-6 ,\
P_2(x,y,z) &= x+2y-7z ,\[2mm]
(C_1)&={ (x,y,z) : S_1(x,y,z)=P_1(x,y,z)=0 } ,\
(C_2)&={ (x,y,z) : S_2(x,y,z)=P_2(x,y,z)=0 } .
end{aligned}
$$
The following solution supports the geometric intuition.
$S_1=0$ describes a sphere centered in $Omega_1=displaystyle left(1,-frac 32,-2right)$ with radius $frac 72$. We cut it with a plane, $P_1=0$, getting a circle.
If this circle is on some other sphere, then this other sphere is centered in a point $Omega$ on the normal form $Omega_1$ to the plane. We already see the plane and its normal, so $Omega$ is on the line with points parametrized as $displaystyle left(1,-frac 32,-2right)+lambda(0,5,6)$, $lambdainBbb R$.
The same game for the other circle.
$S_2=0$ describes a sphere centered in $Omega_2=displaystyle left(frac 32,2, -frac52right)$ with radius $4$. We cut it with a second plane, $P_2=0$, getting a circle.
If this circle is on the same sphere above, then $Omega$ is on the normal form $Omega_2$ to the plane. So $Omega$ is on the line with points parametrized as $displaystyle left(frac 32,2,-frac52right)+mu(1,2,-7)$, $muinBbb R$.
Do these two lines intersect? Yes, $lambda=frac 12$, $mu=-frac 12$, lead to $Omega=(1,1,1)$. So far we land so far as using the hint. ($lambda,mu$ are up to a factor two some $lambda_1,lambda_2$ special values from the hint.)
We still have to check the "radius match". Here we should go back to algebraic geometry.
A point is on $(C_1)$ if it is satisfying for all $lambda_1$ the equation $S_1+lambda_1 P_1=0$.
A point is on $(C_2)$ if it is satisfying for all $lambda_2$ the equation $S_2+lambda_2 P_2=0$.
Now use $lambda_1=2lambda$, $lambda_2=2mu$ (some "equation dedoubling" is the reason for short) and check that we have the same equation. (This corresponds to a "radius match", algebraically to a match of the constant coefficient.)
answered Jan 7 at 18:56
dan_fuleadan_fulea
6,9681413
answered Jan 7 at 18:56
dan_fuleadan_fulea
6,9681413
answered Jan 7 at 18:56
dan_fuleadan_fulea
6,9681413
answered Jan 7 at 18:56
dan_fuleadan_fulea
6,9681413
6,9681413
$begingroup$
Thank you very much sir ! I didn't know this geometrical intuition . But what if the equations of circles are given in their general form like x^2+y^2=r^2 . How can we prove that they lie on same sphere.
$endgroup$
– PN Das
Jan 8 at 15:34
$begingroup$
@PN DAS: The two circles in the problem were given as intersection of a sphere with a plane. For instance, the circle $(C_1)$ is the intersection of the sphere with equation $S_1=0$, with the plane $P_1=0$. Here,$$begin{aligned}S_1(x,y,z) &= x^2 +y^2+z^2 - 2x + 3y + 4z - 5\&=(x-1)^2+(y+3/2)^2+(z+2)^2-49/4end{aligned}$$and from it we extract the center and the radius. A circle cannot be given in 3D by only one equation. An equation like $(x-a)^2+(y-b)^2=r^2$ in 2D, i.e. $(x,y)inBbb R^2$, is a circle, centered in $(a,b)$, radius $r$. In 3D we need one more term² & it's a sphere as above.
$endgroup$
– dan_fulea
Jan 8 at 17:46
$begingroup$
An equation like $x^2+y^2=r^2$ (or more general $(x-a)^2+(y-b)^2=r^2$), seen in three dimensions, i.e. considering the set of all $(x,y,z)$ that satisfy the equation, is a cylinder.
$endgroup$
– dan_fulea
Jan 8 at 17:49
add a comment |
$begingroup$
Thank you very much sir ! I didn't know this geometrical intuition . But what if the equations of circles are given in their general form like x^2+y^2=r^2 . How can we prove that they lie on same sphere.
$endgroup$
– PN Das
Jan 8 at 15:34
$begingroup$
@PN DAS: The two circles in the problem were given as intersection of a sphere with a plane. For instance, the circle $(C_1)$ is the intersection of the sphere with equation $S_1=0$, with the plane $P_1=0$. Here,$$begin{aligned}S_1(x,y,z) &= x^2 +y^2+z^2 - 2x + 3y + 4z - 5\&=(x-1)^2+(y+3/2)^2+(z+2)^2-49/4end{aligned}$$and from it we extract the center and the radius. A circle cannot be given in 3D by only one equation. An equation like $(x-a)^2+(y-b)^2=r^2$ in 2D, i.e. $(x,y)inBbb R^2$, is a circle, centered in $(a,b)$, radius $r$. In 3D we need one more term² & it's a sphere as above.
$endgroup$
– dan_fulea
Jan 8 at 17:46
$begingroup$
An equation like $x^2+y^2=r^2$ (or more general $(x-a)^2+(y-b)^2=r^2$), seen in three dimensions, i.e. considering the set of all $(x,y,z)$ that satisfy the equation, is a cylinder.
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– dan_fulea
Jan 8 at 17:49
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Thank you very much sir ! I didn't know this geometrical intuition . But what if the equations of circles are given in their general form like x^2+y^2=r^2 . How can we prove that they lie on same sphere.
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– PN Das
Jan 8 at 15:34
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Thank you very much sir ! I didn't know this geometrical intuition . But what if the equations of circles are given in their general form like x^2+y^2=r^2 . How can we prove that they lie on same sphere.
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– PN Das
Jan 8 at 15:34
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@PN DAS: The two circles in the problem were given as intersection of a sphere with a plane. For instance, the circle $(C_1)$ is the intersection of the sphere with equation $S_1=0$, with the plane $P_1=0$. Here,$$begin{aligned}S_1(x,y,z) &= x^2 +y^2+z^2 - 2x + 3y + 4z - 5\&=(x-1)^2+(y+3/2)^2+(z+2)^2-49/4end{aligned}$$and from it we extract the center and the radius. A circle cannot be given in 3D by only one equation. An equation like $(x-a)^2+(y-b)^2=r^2$ in 2D, i.e. $(x,y)inBbb R^2$, is a circle, centered in $(a,b)$, radius $r$. In 3D we need one more term² & it's a sphere as above.
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– dan_fulea
Jan 8 at 17:46
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@PN DAS: The two circles in the problem were given as intersection of a sphere with a plane. For instance, the circle $(C_1)$ is the intersection of the sphere with equation $S_1=0$, with the plane $P_1=0$. Here,$$begin{aligned}S_1(x,y,z) &= x^2 +y^2+z^2 - 2x + 3y + 4z - 5\&=(x-1)^2+(y+3/2)^2+(z+2)^2-49/4end{aligned}$$and from it we extract the center and the radius. A circle cannot be given in 3D by only one equation. An equation like $(x-a)^2+(y-b)^2=r^2$ in 2D, i.e. $(x,y)inBbb R^2$, is a circle, centered in $(a,b)$, radius $r$. In 3D we need one more term² & it's a sphere as above.
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– dan_fulea
Jan 8 at 17:46
$begingroup$
An equation like $x^2+y^2=r^2$ (or more general $(x-a)^2+(y-b)^2=r^2$), seen in three dimensions, i.e. considering the set of all $(x,y,z)$ that satisfy the equation, is a cylinder.
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– dan_fulea
Jan 8 at 17:49
$begingroup$
An equation like $x^2+y^2=r^2$ (or more general $(x-a)^2+(y-b)^2=r^2$), seen in three dimensions, i.e. considering the set of all $(x,y,z)$ that satisfy the equation, is a cylinder.
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– dan_fulea
Jan 8 at 17:49
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It might help to give the title and author of your textbook. By doing only the second part you seem to mean finding the equation of the sphere. If you have done that, doesn't it become fairly straightforward to check that the circles lie on the sphere by comparing the coordinates?
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– hardmath
Jan 9 at 3:17