Condition of conic to be a circle

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Let $a,b,h,g,f,c in mathbbR$. Then the general equation of a conic given by:
$ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ represents the equation of a circle iff
beginalign
beginvmatrix
a & h & g \
h & b & f \
g & f & c
endvmatrix leq 0, h = 0, a = b neq 0.
endalign
For the $Leftarrow$, it is easy to see as if you plug in the values, it can be easily written as $(x + fracga)^2 + (y + fracfa)^2 = frac1a^2(g^2 + f^2 - ac) $ which is a circle. I am having trouble with the $Rightarrow$ part.

I am using the following definition of a circle:

A circle is a set of points $(x,y) in mathbbR^2$ which satisfies the equation:
$(x- X)^2 + (y - Y)^2 = R^2$ for some $(X,Y) in mathbbR^2$, $R in mathbbR_geq 0$.

Using this definition of a circle, how do I show the $Rightarrow$ part of the above result for the general conic to be a circle?

circle

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asked Jul 17 at 6:40

Ashish

304211

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Let $a,b,h,g,f,c in mathbbR$. Then the general equation of a conic given by:
$ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ represents the equation of a circle iff
beginalign
beginvmatrix
a & h & g \
h & b & f \
g & f & c
endvmatrix leq 0, h = 0, a = b neq 0.
endalign
For the $Leftarrow$, it is easy to see as if you plug in the values, it can be easily written as $(x + fracga)^2 + (y + fracfa)^2 = frac1a^2(g^2 + f^2 - ac) $ which is a circle. I am having trouble with the $Rightarrow$ part.

I am using the following definition of a circle:

A circle is a set of points $(x,y) in mathbbR^2$ which satisfies the equation:
$(x- X)^2 + (y - Y)^2 = R^2$ for some $(X,Y) in mathbbR^2$, $R in mathbbR_geq 0$.

Using this definition of a circle, how do I show the $Rightarrow$ part of the above result for the general conic to be a circle?

circle

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asked Jul 17 at 6:40

Ashish

304211

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $a,b,h,g,f,c in mathbbR$. Then the general equation of a conic given by:
$ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ represents the equation of a circle iff
beginalign
beginvmatrix
a & h & g \
h & b & f \
g & f & c
endvmatrix leq 0, h = 0, a = b neq 0.
endalign
For the $Leftarrow$, it is easy to see as if you plug in the values, it can be easily written as $(x + fracga)^2 + (y + fracfa)^2 = frac1a^2(g^2 + f^2 - ac) $ which is a circle. I am having trouble with the $Rightarrow$ part.

I am using the following definition of a circle:

A circle is a set of points $(x,y) in mathbbR^2$ which satisfies the equation:
$(x- X)^2 + (y - Y)^2 = R^2$ for some $(X,Y) in mathbbR^2$, $R in mathbbR_geq 0$.

Using this definition of a circle, how do I show the $Rightarrow$ part of the above result for the general conic to be a circle?

circle

share|cite|improve this question

asked Jul 17 at 6:40

Ashish

304211

Let $a,b,h,g,f,c in mathbbR$. Then the general equation of a conic given by:
$ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ represents the equation of a circle iff
beginalign
beginvmatrix
a & h & g \
h & b & f \
g & f & c
endvmatrix leq 0, h = 0, a = b neq 0.
endalign
For the $Leftarrow$, it is easy to see as if you plug in the values, it can be easily written as $(x + fracga)^2 + (y + fracfa)^2 = frac1a^2(g^2 + f^2 - ac) $ which is a circle. I am having trouble with the $Rightarrow$ part.

I am using the following definition of a circle:

A circle is a set of points $(x,y) in mathbbR^2$ which satisfies the equation:
$(x- X)^2 + (y - Y)^2 = R^2$ for some $(X,Y) in mathbbR^2$, $R in mathbbR_geq 0$.

Using this definition of a circle, how do I show the $Rightarrow$ part of the above result for the general conic to be a circle?

circle

share|cite|improve this question

asked Jul 17 at 6:40

Ashish

304211

share|cite|improve this question

share|cite|improve this question

asked Jul 17 at 6:40

Ashish

304211

asked Jul 17 at 6:40

Ashish

304211

asked Jul 17 at 6:40

Ashish

304211

304211

add a comment | 

add a comment | 

2 Answers
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@amd: Since my comment was too big, I had to post it as an answer.

$S subseteq mathbbR^2$ is defined as a circle if $exists X,Y in mathbbR$, $R in mathbbR_geq 0$ such that $(x - X)^2 + (y - Y)^2 = R^2 forall (x,y) in S$.

'$Leftarrow$' proof:

If it is given $ a = b neq 0$, $h = 0$, $beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix leq 0$ then $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ can be rewritten as $(x + fracga)^2 + (y + fracfb)^2 = frac1a^2(g^2 + f^2 - ac)$. So I can give (X,Y) = $(-fracga,-fracfb)$ and $R = frac1asqrtg^2 + f^2 - ac geq 0$ such that every $(x,y)$ which satisfies the locus $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ satisfies $(x-X)^2 + (y-Y)^2 = R^2 $. Hence the set of points which satisfies $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ is a circle.

'$Rightarrow$' proof:

It is given that $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ represents a circle. That is $exists X,Y in mathbbR$, $R in mathbbR_geq 0$ such that $(x - X)^2 + (y - Y)^2 = R^2$ for all $(x,y) $ satisfying the equation $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$. Now how do I proceed from here?

If I am supposed to match the terms, what is the formal reasoning to do so? Agreed that $x^2,y^2 ... $ etc are LI terms, and then how to go about from there?

share|cite|improve this answer

answered Jul 18 at 9:27

Ashish

304211

add a comment | 

up vote
-1
down vote

We have
$$(x-X)^2 + (y-Y)^2 = R^2$$
$$Leftrightarrow$$
$$x^2 + y^2 - 2Xx - 2Yy + X^2 + Y^2 - R^2 = 0,$$
and matching terms with
$$ ax^2+bx^2+hxy+2gx+2fy+c = 0 $$
gives
$$left{
beginarrayl
a = 1\
b = 1\
h = 0\
g = -X\
f = -Y\
c = X^2+Y^2-R^2
endarray
right..$$
Hence, $,a = b not= 0,, h=0,$ and
beginalign
beginvmatrix
a & h & g\
h & b & f\
g & f & c
endvmatrix
=
beginvmatrix
1 & 0 & -X\
0 & 1 & -Y\
-X & -Y & X^2+Y^2-R^2
endvmatrix
= -R^2 leq 0.
endalign

share|cite|improve this answer

answered Jul 17 at 7:03

Sobi

558310

• 
  
  
  

  
  

  
  
  
  
  
  

  
  No no; I've seen this matching terms argument before.. but my problem with it is that it's not a formal reasoning i.e. on what basis can you start matching terms? Or can you prove that if two equations represent the same locus of points, then their terms can be matched? And btw, this isn't correct as you have shown that if a general conic is a circle, then $a = b = 1$ which isn't correct, as you can have $a = b = 2$ and it will still be a circle.
  – Ashish
  Jul 17 at 7:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The factors $x^2, y^2,$ etc are linearly independent, so you can match terms. As for your second question, it does not matter if you multiply the whole equations by a constant factor, and your other conditions remain unchanged then as well. In particular, if you have $a=b=2$ in your equation, you can divide it by 2 to arrive at my computations.
  – Sobi
  Jul 17 at 12:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ashish You’re effectively doing the same sort of term matching in your own proof of the implication in the other direction, but you’ve got no objection to it there.
  – amd
  Jul 18 at 2:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sobi Could you formalize the argument? i.e. while incorporating the constant factor multiplication to the whole equation doesn't change it's locus in the reasoning? I'm completely confused.
  – Ashish
  Jul 18 at 5:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Amd. No I'm not. I refactored to write the equation $ax^2 + ... = 0$ in the form of $(x - X)^2 + (y - Y)^2 = R^2$ where $(X,Y) = (-fracga, -fracfa)$ and $R = ...$, which is in accordance to the 'definition' of circle.
  – Ashish
  Jul 18 at 5:02
  
  

  
  
  
  

 | 
show 2 more comments

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2 Answers
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2 Answers
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active

oldest

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up vote
0
down vote

@amd: Since my comment was too big, I had to post it as an answer.

$S subseteq mathbbR^2$ is defined as a circle if $exists X,Y in mathbbR$, $R in mathbbR_geq 0$ such that $(x - X)^2 + (y - Y)^2 = R^2 forall (x,y) in S$.

'$Leftarrow$' proof:

If it is given $ a = b neq 0$, $h = 0$, $beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix leq 0$ then $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ can be rewritten as $(x + fracga)^2 + (y + fracfb)^2 = frac1a^2(g^2 + f^2 - ac)$. So I can give (X,Y) = $(-fracga,-fracfb)$ and $R = frac1asqrtg^2 + f^2 - ac geq 0$ such that every $(x,y)$ which satisfies the locus $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ satisfies $(x-X)^2 + (y-Y)^2 = R^2 $. Hence the set of points which satisfies $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ is a circle.

'$Rightarrow$' proof:

It is given that $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ represents a circle. That is $exists X,Y in mathbbR$, $R in mathbbR_geq 0$ such that $(x - X)^2 + (y - Y)^2 = R^2$ for all $(x,y) $ satisfying the equation $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$. Now how do I proceed from here?

If I am supposed to match the terms, what is the formal reasoning to do so? Agreed that $x^2,y^2 ... $ etc are LI terms, and then how to go about from there?

share|cite|improve this answer

answered Jul 18 at 9:27

Ashish

304211

add a comment | 

up vote
0
down vote

@amd: Since my comment was too big, I had to post it as an answer.

$S subseteq mathbbR^2$ is defined as a circle if $exists X,Y in mathbbR$, $R in mathbbR_geq 0$ such that $(x - X)^2 + (y - Y)^2 = R^2 forall (x,y) in S$.

'$Leftarrow$' proof:

If it is given $ a = b neq 0$, $h = 0$, $beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix leq 0$ then $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ can be rewritten as $(x + fracga)^2 + (y + fracfb)^2 = frac1a^2(g^2 + f^2 - ac)$. So I can give (X,Y) = $(-fracga,-fracfb)$ and $R = frac1asqrtg^2 + f^2 - ac geq 0$ such that every $(x,y)$ which satisfies the locus $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ satisfies $(x-X)^2 + (y-Y)^2 = R^2 $. Hence the set of points which satisfies $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ is a circle.

'$Rightarrow$' proof:

It is given that $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ represents a circle. That is $exists X,Y in mathbbR$, $R in mathbbR_geq 0$ such that $(x - X)^2 + (y - Y)^2 = R^2$ for all $(x,y) $ satisfying the equation $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$. Now how do I proceed from here?

If I am supposed to match the terms, what is the formal reasoning to do so? Agreed that $x^2,y^2 ... $ etc are LI terms, and then how to go about from there?

share|cite|improve this answer

answered Jul 18 at 9:27

Ashish

304211

add a comment | 

up vote
0
down vote

up vote
0
down vote

@amd: Since my comment was too big, I had to post it as an answer.

$S subseteq mathbbR^2$ is defined as a circle if $exists X,Y in mathbbR$, $R in mathbbR_geq 0$ such that $(x - X)^2 + (y - Y)^2 = R^2 forall (x,y) in S$.

'$Leftarrow$' proof:

If it is given $ a = b neq 0$, $h = 0$, $beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix leq 0$ then $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ can be rewritten as $(x + fracga)^2 + (y + fracfb)^2 = frac1a^2(g^2 + f^2 - ac)$. So I can give (X,Y) = $(-fracga,-fracfb)$ and $R = frac1asqrtg^2 + f^2 - ac geq 0$ such that every $(x,y)$ which satisfies the locus $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ satisfies $(x-X)^2 + (y-Y)^2 = R^2 $. Hence the set of points which satisfies $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ is a circle.

'$Rightarrow$' proof:

It is given that $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ represents a circle. That is $exists X,Y in mathbbR$, $R in mathbbR_geq 0$ such that $(x - X)^2 + (y - Y)^2 = R^2$ for all $(x,y) $ satisfying the equation $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$. Now how do I proceed from here?

If I am supposed to match the terms, what is the formal reasoning to do so? Agreed that $x^2,y^2 ... $ etc are LI terms, and then how to go about from there?

share|cite|improve this answer

answered Jul 18 at 9:27

Ashish

304211

@amd: Since my comment was too big, I had to post it as an answer.

$S subseteq mathbbR^2$ is defined as a circle if $exists X,Y in mathbbR$, $R in mathbbR_geq 0$ such that $(x - X)^2 + (y - Y)^2 = R^2 forall (x,y) in S$.

'$Leftarrow$' proof:

If it is given $ a = b neq 0$, $h = 0$, $beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix leq 0$ then $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ can be rewritten as $(x + fracga)^2 + (y + fracfb)^2 = frac1a^2(g^2 + f^2 - ac)$. So I can give (X,Y) = $(-fracga,-fracfb)$ and $R = frac1asqrtg^2 + f^2 - ac geq 0$ such that every $(x,y)$ which satisfies the locus $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ satisfies $(x-X)^2 + (y-Y)^2 = R^2 $. Hence the set of points which satisfies $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ is a circle.

'$Rightarrow$' proof:

It is given that $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$ represents a circle. That is $exists X,Y in mathbbR$, $R in mathbbR_geq 0$ such that $(x - X)^2 + (y - Y)^2 = R^2$ for all $(x,y) $ satisfying the equation $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$. Now how do I proceed from here?

If I am supposed to match the terms, what is the formal reasoning to do so? Agreed that $x^2,y^2 ... $ etc are LI terms, and then how to go about from there?

share|cite|improve this answer

answered Jul 18 at 9:27

Ashish

304211

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 9:27

Ashish

304211

answered Jul 18 at 9:27

Ashish

304211

answered Jul 18 at 9:27

Ashish

304211

304211

add a comment | 

add a comment | 

up vote
-1
down vote

We have
$$(x-X)^2 + (y-Y)^2 = R^2$$
$$Leftrightarrow$$
$$x^2 + y^2 - 2Xx - 2Yy + X^2 + Y^2 - R^2 = 0,$$
and matching terms with
$$ ax^2+bx^2+hxy+2gx+2fy+c = 0 $$
gives
$$left{
beginarrayl
a = 1\
b = 1\
h = 0\
g = -X\
f = -Y\
c = X^2+Y^2-R^2
endarray
right..$$
Hence, $,a = b not= 0,, h=0,$ and
beginalign
beginvmatrix
a & h & g\
h & b & f\
g & f & c
endvmatrix
=
beginvmatrix
1 & 0 & -X\
0 & 1 & -Y\
-X & -Y & X^2+Y^2-R^2
endvmatrix
= -R^2 leq 0.
endalign

share|cite|improve this answer

answered Jul 17 at 7:03

Sobi

558310

• 
  
  
  

  
  

  
  
  
  
  
  

  
  No no; I've seen this matching terms argument before.. but my problem with it is that it's not a formal reasoning i.e. on what basis can you start matching terms? Or can you prove that if two equations represent the same locus of points, then their terms can be matched? And btw, this isn't correct as you have shown that if a general conic is a circle, then $a = b = 1$ which isn't correct, as you can have $a = b = 2$ and it will still be a circle.
  – Ashish
  Jul 17 at 7:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The factors $x^2, y^2,$ etc are linearly independent, so you can match terms. As for your second question, it does not matter if you multiply the whole equations by a constant factor, and your other conditions remain unchanged then as well. In particular, if you have $a=b=2$ in your equation, you can divide it by 2 to arrive at my computations.
  – Sobi
  Jul 17 at 12:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ashish You’re effectively doing the same sort of term matching in your own proof of the implication in the other direction, but you’ve got no objection to it there.
  – amd
  Jul 18 at 2:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sobi Could you formalize the argument? i.e. while incorporating the constant factor multiplication to the whole equation doesn't change it's locus in the reasoning? I'm completely confused.
  – Ashish
  Jul 18 at 5:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Amd. No I'm not. I refactored to write the equation $ax^2 + ... = 0$ in the form of $(x - X)^2 + (y - Y)^2 = R^2$ where $(X,Y) = (-fracga, -fracfa)$ and $R = ...$, which is in accordance to the 'definition' of circle.
  – Ashish
  Jul 18 at 5:02
  
  

  
  
  
  

 | 
show 2 more comments

up vote
-1
down vote

We have
$$(x-X)^2 + (y-Y)^2 = R^2$$
$$Leftrightarrow$$
$$x^2 + y^2 - 2Xx - 2Yy + X^2 + Y^2 - R^2 = 0,$$
and matching terms with
$$ ax^2+bx^2+hxy+2gx+2fy+c = 0 $$
gives
$$left{
beginarrayl
a = 1\
b = 1\
h = 0\
g = -X\
f = -Y\
c = X^2+Y^2-R^2
endarray
right..$$
Hence, $,a = b not= 0,, h=0,$ and
beginalign
beginvmatrix
a & h & g\
h & b & f\
g & f & c
endvmatrix
=
beginvmatrix
1 & 0 & -X\
0 & 1 & -Y\
-X & -Y & X^2+Y^2-R^2
endvmatrix
= -R^2 leq 0.
endalign

share|cite|improve this answer

answered Jul 17 at 7:03

Sobi

558310

• 
  
  
  

  
  

  
  
  
  
  
  

  
  No no; I've seen this matching terms argument before.. but my problem with it is that it's not a formal reasoning i.e. on what basis can you start matching terms? Or can you prove that if two equations represent the same locus of points, then their terms can be matched? And btw, this isn't correct as you have shown that if a general conic is a circle, then $a = b = 1$ which isn't correct, as you can have $a = b = 2$ and it will still be a circle.
  – Ashish
  Jul 17 at 7:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The factors $x^2, y^2,$ etc are linearly independent, so you can match terms. As for your second question, it does not matter if you multiply the whole equations by a constant factor, and your other conditions remain unchanged then as well. In particular, if you have $a=b=2$ in your equation, you can divide it by 2 to arrive at my computations.
  – Sobi
  Jul 17 at 12:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ashish You’re effectively doing the same sort of term matching in your own proof of the implication in the other direction, but you’ve got no objection to it there.
  – amd
  Jul 18 at 2:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sobi Could you formalize the argument? i.e. while incorporating the constant factor multiplication to the whole equation doesn't change it's locus in the reasoning? I'm completely confused.
  – Ashish
  Jul 18 at 5:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Amd. No I'm not. I refactored to write the equation $ax^2 + ... = 0$ in the form of $(x - X)^2 + (y - Y)^2 = R^2$ where $(X,Y) = (-fracga, -fracfa)$ and $R = ...$, which is in accordance to the 'definition' of circle.
  – Ashish
  Jul 18 at 5:02
  
  

  
  
  
  

 | 
show 2 more comments

up vote
-1
down vote

up vote
-1
down vote

We have
$$(x-X)^2 + (y-Y)^2 = R^2$$
$$Leftrightarrow$$
$$x^2 + y^2 - 2Xx - 2Yy + X^2 + Y^2 - R^2 = 0,$$
and matching terms with
$$ ax^2+bx^2+hxy+2gx+2fy+c = 0 $$
gives
$$left{
beginarrayl
a = 1\
b = 1\
h = 0\
g = -X\
f = -Y\
c = X^2+Y^2-R^2
endarray
right..$$
Hence, $,a = b not= 0,, h=0,$ and
beginalign
beginvmatrix
a & h & g\
h & b & f\
g & f & c
endvmatrix
=
beginvmatrix
1 & 0 & -X\
0 & 1 & -Y\
-X & -Y & X^2+Y^2-R^2
endvmatrix
= -R^2 leq 0.
endalign

share|cite|improve this answer

answered Jul 17 at 7:03

Sobi

558310

We have
$$(x-X)^2 + (y-Y)^2 = R^2$$
$$Leftrightarrow$$
$$x^2 + y^2 - 2Xx - 2Yy + X^2 + Y^2 - R^2 = 0,$$
and matching terms with
$$ ax^2+bx^2+hxy+2gx+2fy+c = 0 $$
gives
$$left{
beginarrayl
a = 1\
b = 1\
h = 0\
g = -X\
f = -Y\
c = X^2+Y^2-R^2
endarray
right..$$
Hence, $,a = b not= 0,, h=0,$ and
beginalign
beginvmatrix
a & h & g\
h & b & f\
g & f & c
endvmatrix
=
beginvmatrix
1 & 0 & -X\
0 & 1 & -Y\
-X & -Y & X^2+Y^2-R^2
endvmatrix
= -R^2 leq 0.
endalign

share|cite|improve this answer

answered Jul 17 at 7:03

Sobi

558310

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 7:03

Sobi

558310

answered Jul 17 at 7:03

Sobi

558310

answered Jul 17 at 7:03

Sobi

558310

558310

• 
  
  
  

  
  

  
  
  
  
  
  

  
  No no; I've seen this matching terms argument before.. but my problem with it is that it's not a formal reasoning i.e. on what basis can you start matching terms? Or can you prove that if two equations represent the same locus of points, then their terms can be matched? And btw, this isn't correct as you have shown that if a general conic is a circle, then $a = b = 1$ which isn't correct, as you can have $a = b = 2$ and it will still be a circle.
  – Ashish
  Jul 17 at 7:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The factors $x^2, y^2,$ etc are linearly independent, so you can match terms. As for your second question, it does not matter if you multiply the whole equations by a constant factor, and your other conditions remain unchanged then as well. In particular, if you have $a=b=2$ in your equation, you can divide it by 2 to arrive at my computations.
  – Sobi
  Jul 17 at 12:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ashish You’re effectively doing the same sort of term matching in your own proof of the implication in the other direction, but you’ve got no objection to it there.
  – amd
  Jul 18 at 2:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sobi Could you formalize the argument? i.e. while incorporating the constant factor multiplication to the whole equation doesn't change it's locus in the reasoning? I'm completely confused.
  – Ashish
  Jul 18 at 5:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Amd. No I'm not. I refactored to write the equation $ax^2 + ... = 0$ in the form of $(x - X)^2 + (y - Y)^2 = R^2$ where $(X,Y) = (-fracga, -fracfa)$ and $R = ...$, which is in accordance to the 'definition' of circle.
  – Ashish
  Jul 18 at 5:02
  
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  No no; I've seen this matching terms argument before.. but my problem with it is that it's not a formal reasoning i.e. on what basis can you start matching terms? Or can you prove that if two equations represent the same locus of points, then their terms can be matched? And btw, this isn't correct as you have shown that if a general conic is a circle, then $a = b = 1$ which isn't correct, as you can have $a = b = 2$ and it will still be a circle.
  – Ashish
  Jul 17 at 7:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The factors $x^2, y^2,$ etc are linearly independent, so you can match terms. As for your second question, it does not matter if you multiply the whole equations by a constant factor, and your other conditions remain unchanged then as well. In particular, if you have $a=b=2$ in your equation, you can divide it by 2 to arrive at my computations.
  – Sobi
  Jul 17 at 12:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ashish You’re effectively doing the same sort of term matching in your own proof of the implication in the other direction, but you’ve got no objection to it there.
  – amd
  Jul 18 at 2:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sobi Could you formalize the argument? i.e. while incorporating the constant factor multiplication to the whole equation doesn't change it's locus in the reasoning? I'm completely confused.
  – Ashish
  Jul 18 at 5:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Amd. No I'm not. I refactored to write the equation $ax^2 + ... = 0$ in the form of $(x - X)^2 + (y - Y)^2 = R^2$ where $(X,Y) = (-fracga, -fracfa)$ and $R = ...$, which is in accordance to the 'definition' of circle.
  – Ashish
  Jul 18 at 5:02
  
  

  
  
  
  

No no; I've seen this matching terms argument before.. but my problem with it is that it's not a formal reasoning i.e. on what basis can you start matching terms? Or can you prove that if two equations represent the same locus of points, then their terms can be matched? And btw, this isn't correct as you have shown that if a general conic is a circle, then $a = b = 1$ which isn't correct, as you can have $a = b = 2$ and it will still be a circle.
– Ashish
Jul 17 at 7:47

No no; I've seen this matching terms argument before.. but my problem with it is that it's not a formal reasoning i.e. on what basis can you start matching terms? Or can you prove that if two equations represent the same locus of points, then their terms can be matched? And btw, this isn't correct as you have shown that if a general conic is a circle, then $a = b = 1$ which isn't correct, as you can have $a = b = 2$ and it will still be a circle.
– Ashish
Jul 17 at 7:47

The factors $x^2, y^2,$ etc are linearly independent, so you can match terms. As for your second question, it does not matter if you multiply the whole equations by a constant factor, and your other conditions remain unchanged then as well. In particular, if you have $a=b=2$ in your equation, you can divide it by 2 to arrive at my computations.
– Sobi
Jul 17 at 12:43

The factors $x^2, y^2,$ etc are linearly independent, so you can match terms. As for your second question, it does not matter if you multiply the whole equations by a constant factor, and your other conditions remain unchanged then as well. In particular, if you have $a=b=2$ in your equation, you can divide it by 2 to arrive at my computations.
– Sobi
Jul 17 at 12:43

@Ashish You’re effectively doing the same sort of term matching in your own proof of the implication in the other direction, but you’ve got no objection to it there.
– amd
Jul 18 at 2:16

@Ashish You’re effectively doing the same sort of term matching in your own proof of the implication in the other direction, but you’ve got no objection to it there.
– amd
Jul 18 at 2:16

@Sobi Could you formalize the argument? i.e. while incorporating the constant factor multiplication to the whole equation doesn't change it's locus in the reasoning? I'm completely confused.
– Ashish
Jul 18 at 5:02

@Sobi Could you formalize the argument? i.e. while incorporating the constant factor multiplication to the whole equation doesn't change it's locus in the reasoning? I'm completely confused.
– Ashish
Jul 18 at 5:02

@Amd. No I'm not. I refactored to write the equation $ax^2 + ... = 0$ in the form of $(x - X)^2 + (y - Y)^2 = R^2$ where $(X,Y) = (-fracga, -fracfa)$ and $R = ...$, which is in accordance to the 'definition' of circle.
– Ashish
Jul 18 at 5:02

@Amd. No I'm not. I refactored to write the equation $ax^2 + ... = 0$ in the form of $(x - X)^2 + (y - Y)^2 = R^2$ where $(X,Y) = (-fracga, -fracfa)$ and $R = ...$, which is in accordance to the 'definition' of circle.
– Ashish
Jul 18 at 5:02

 | 
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