Condition for which $|z-z_1|^2+|z-z_2|^2=Kinmathbb{R}$ represent a circle
$begingroup$
Let $Kinmathbb{R}$ and $z_1,z_2inmathbb{C}$.
Prove that the equation $|z-z_1|^2+|z-z_2|^2=K$ represent a circle iff $Kgeqfrac{1}{2}|z_1-z_2|^2$.
My Attempt
$$
|z|^2+|z_1|^2-2mathcal{Re}(zbar{z}_1)+|z|^2+|z_2|^2-2mathcal{Re}(zbar{z}_2)=K\
2|z|^2+|z_1|^2+|z_2|^2-2bigg[mathcal{Re}(zbar{z}_1)+mathcal{Re}(zbar{z}_2)bigg]=K\
2|z|^2+|z_1|^2+|z_2|^2-2bigg[ a_1x+b_1y+a_2x+b_2y bigg]=K\
2|z|^2+|z_1|^2+|z_2|^2-2mathcal{Re}bigg[ zbar{z}_1+zbar{z}_2 bigg]=K\
2|z|^2+|z_1|^2+|z_2|^2-2mathcal{Re}bigg[ z(bar{z}_1+bar{z}_2) bigg]=K\
|z|^2+frac{1}{2}(|z_1|^2+|z_2|^2)-2mathcal{Re}Big(z.frac{bar{z}_1+bar{z}_2}{2}Big)=frac{K}{2}\
|z|^2+frac{1}{2}(|z_1|^2+|z_2|^2)+Big|frac{z_1+z_2}{2}Big|^2-2mathcal{Re}Big(z.frac{bar{z}_1+bar{z}_2}{2}Big)=frac{K}{2}+Big|frac{z_1+z_2}{2}Big|^2\
|z|^2+big|frac{z_1+z_2}{2}big|^2-2mathcal{Re}Big(z.frac{bar{z}_1+bar{z}_2}{2}Big)=frac{K}{2}+Big|frac{z_1+z_2}{2}Big|^2-frac{1}{2}(|z_1|^2+|z_2|^2)\
Big|z-frac{z_1+z_2}{2}Big|^2=frac{K}{2}+frac{1}{2}(z_1bar{z}_2+bar{z}_1z_2)=frac{K}{2}+frac{|z_1|^2+|z_2|^2}{4}+frac{z_1bar{z}_2+bar{z}_1z_2}{4}-frac{1}{2}(z_1bar{z}_2+bar{z}_1z_2)\
Big|z-frac{z_1+z_2}{2}Big|^2=frac{K}{2}+frac{1}{4}(|z_1|^2+|z_2|^2-z_1bar{z}_2-bar{z}_1z_2)=frac{K}{2}-frac{1}{4}|z_1-z_2|^2\
Big|z-frac{z_1+z_2}{2}Big|^2=frac{1}{4}Big[2K-|z_1-z_2|^2Big]\
implies 2Kgeq|z_1-z_2|^2implies boxed{Kgeqfrac{1}{2}|z_1-z_2|^2}
$$
As it was asked as a multiple choice question I think my attempt is a bit long and cumbersome. So, is there an easier way to solve this ?
complex-numbers vectors circle
asked Jan 17 at 11:13
ss1729ss1729
1,9331923
$endgroup$
add a comment |
$begingroup$
Let $Kinmathbb{R}$ and $z_1,z_2inmathbb{C}$.
Prove that the equation $|z-z_1|^2+|z-z_2|^2=K$ represent a circle iff $Kgeqfrac{1}{2}|z_1-z_2|^2$.
My Attempt
$$
|z|^2+|z_1|^2-2mathcal{Re}(zbar{z}_1)+|z|^2+|z_2|^2-2mathcal{Re}(zbar{z}_2)=K\
2|z|^2+|z_1|^2+|z_2|^2-2bigg[mathcal{Re}(zbar{z}_1)+mathcal{Re}(zbar{z}_2)bigg]=K\
2|z|^2+|z_1|^2+|z_2|^2-2bigg[ a_1x+b_1y+a_2x+b_2y bigg]=K\
2|z|^2+|z_1|^2+|z_2|^2-2mathcal{Re}bigg[ zbar{z}_1+zbar{z}_2 bigg]=K\
2|z|^2+|z_1|^2+|z_2|^2-2mathcal{Re}bigg[ z(bar{z}_1+bar{z}_2) bigg]=K\
|z|^2+frac{1}{2}(|z_1|^2+|z_2|^2)-2mathcal{Re}Big(z.frac{bar{z}_1+bar{z}_2}{2}Big)=frac{K}{2}\
|z|^2+frac{1}{2}(|z_1|^2+|z_2|^2)+Big|frac{z_1+z_2}{2}Big|^2-2mathcal{Re}Big(z.frac{bar{z}_1+bar{z}_2}{2}Big)=frac{K}{2}+Big|frac{z_1+z_2}{2}Big|^2\
|z|^2+big|frac{z_1+z_2}{2}big|^2-2mathcal{Re}Big(z.frac{bar{z}_1+bar{z}_2}{2}Big)=frac{K}{2}+Big|frac{z_1+z_2}{2}Big|^2-frac{1}{2}(|z_1|^2+|z_2|^2)\
Big|z-frac{z_1+z_2}{2}Big|^2=frac{K}{2}+frac{1}{2}(z_1bar{z}_2+bar{z}_1z_2)=frac{K}{2}+frac{|z_1|^2+|z_2|^2}{4}+frac{z_1bar{z}_2+bar{z}_1z_2}{4}-frac{1}{2}(z_1bar{z}_2+bar{z}_1z_2)\
Big|z-frac{z_1+z_2}{2}Big|^2=frac{K}{2}+frac{1}{4}(|z_1|^2+|z_2|^2-z_1bar{z}_2-bar{z}_1z_2)=frac{K}{2}-frac{1}{4}|z_1-z_2|^2\
Big|z-frac{z_1+z_2}{2}Big|^2=frac{1}{4}Big[2K-|z_1-z_2|^2Big]\
implies 2Kgeq|z_1-z_2|^2implies boxed{Kgeqfrac{1}{2}|z_1-z_2|^2}
$$
As it was asked as a multiple choice question I think my attempt is a bit long and cumbersome. So, is there an easier way to solve this ?
complex-numbers vectors circle
asked Jan 17 at 11:13
ss1729ss1729
1,9331923
$endgroup$
add a comment |
$begingroup$
Let $Kinmathbb{R}$ and $z_1,z_2inmathbb{C}$.
Prove that the equation $|z-z_1|^2+|z-z_2|^2=K$ represent a circle iff $Kgeqfrac{1}{2}|z_1-z_2|^2$.
My Attempt
$$
|z|^2+|z_1|^2-2mathcal{Re}(zbar{z}_1)+|z|^2+|z_2|^2-2mathcal{Re}(zbar{z}_2)=K\
2|z|^2+|z_1|^2+|z_2|^2-2bigg[mathcal{Re}(zbar{z}_1)+mathcal{Re}(zbar{z}_2)bigg]=K\
2|z|^2+|z_1|^2+|z_2|^2-2bigg[ a_1x+b_1y+a_2x+b_2y bigg]=K\
2|z|^2+|z_1|^2+|z_2|^2-2mathcal{Re}bigg[ zbar{z}_1+zbar{z}_2 bigg]=K\
2|z|^2+|z_1|^2+|z_2|^2-2mathcal{Re}bigg[ z(bar{z}_1+bar{z}_2) bigg]=K\
|z|^2+frac{1}{2}(|z_1|^2+|z_2|^2)-2mathcal{Re}Big(z.frac{bar{z}_1+bar{z}_2}{2}Big)=frac{K}{2}\
|z|^2+frac{1}{2}(|z_1|^2+|z_2|^2)+Big|frac{z_1+z_2}{2}Big|^2-2mathcal{Re}Big(z.frac{bar{z}_1+bar{z}_2}{2}Big)=frac{K}{2}+Big|frac{z_1+z_2}{2}Big|^2\
|z|^2+big|frac{z_1+z_2}{2}big|^2-2mathcal{Re}Big(z.frac{bar{z}_1+bar{z}_2}{2}Big)=frac{K}{2}+Big|frac{z_1+z_2}{2}Big|^2-frac{1}{2}(|z_1|^2+|z_2|^2)\
Big|z-frac{z_1+z_2}{2}Big|^2=frac{K}{2}+frac{1}{2}(z_1bar{z}_2+bar{z}_1z_2)=frac{K}{2}+frac{|z_1|^2+|z_2|^2}{4}+frac{z_1bar{z}_2+bar{z}_1z_2}{4}-frac{1}{2}(z_1bar{z}_2+bar{z}_1z_2)\
Big|z-frac{z_1+z_2}{2}Big|^2=frac{K}{2}+frac{1}{4}(|z_1|^2+|z_2|^2-z_1bar{z}_2-bar{z}_1z_2)=frac{K}{2}-frac{1}{4}|z_1-z_2|^2\
Big|z-frac{z_1+z_2}{2}Big|^2=frac{1}{4}Big[2K-|z_1-z_2|^2Big]\
implies 2Kgeq|z_1-z_2|^2implies boxed{Kgeqfrac{1}{2}|z_1-z_2|^2}
$$
As it was asked as a multiple choice question I think my attempt is a bit long and cumbersome. So, is there an easier way to solve this ?
complex-numbers vectors circle
asked Jan 17 at 11:13
ss1729ss1729
1,9331923
$endgroup$
Let $Kinmathbb{R}$ and $z_1,z_2inmathbb{C}$.
Prove that the equation $|z-z_1|^2+|z-z_2|^2=K$ represent a circle iff $Kgeqfrac{1}{2}|z_1-z_2|^2$.
My Attempt
$$
|z|^2+|z_1|^2-2mathcal{Re}(zbar{z}_1)+|z|^2+|z_2|^2-2mathcal{Re}(zbar{z}_2)=K\
2|z|^2+|z_1|^2+|z_2|^2-2bigg[mathcal{Re}(zbar{z}_1)+mathcal{Re}(zbar{z}_2)bigg]=K\
2|z|^2+|z_1|^2+|z_2|^2-2bigg[ a_1x+b_1y+a_2x+b_2y bigg]=K\
2|z|^2+|z_1|^2+|z_2|^2-2mathcal{Re}bigg[ zbar{z}_1+zbar{z}_2 bigg]=K\
2|z|^2+|z_1|^2+|z_2|^2-2mathcal{Re}bigg[ z(bar{z}_1+bar{z}_2) bigg]=K\
|z|^2+frac{1}{2}(|z_1|^2+|z_2|^2)-2mathcal{Re}Big(z.frac{bar{z}_1+bar{z}_2}{2}Big)=frac{K}{2}\
|z|^2+frac{1}{2}(|z_1|^2+|z_2|^2)+Big|frac{z_1+z_2}{2}Big|^2-2mathcal{Re}Big(z.frac{bar{z}_1+bar{z}_2}{2}Big)=frac{K}{2}+Big|frac{z_1+z_2}{2}Big|^2\
|z|^2+big|frac{z_1+z_2}{2}big|^2-2mathcal{Re}Big(z.frac{bar{z}_1+bar{z}_2}{2}Big)=frac{K}{2}+Big|frac{z_1+z_2}{2}Big|^2-frac{1}{2}(|z_1|^2+|z_2|^2)\
Big|z-frac{z_1+z_2}{2}Big|^2=frac{K}{2}+frac{1}{2}(z_1bar{z}_2+bar{z}_1z_2)=frac{K}{2}+frac{|z_1|^2+|z_2|^2}{4}+frac{z_1bar{z}_2+bar{z}_1z_2}{4}-frac{1}{2}(z_1bar{z}_2+bar{z}_1z_2)\
Big|z-frac{z_1+z_2}{2}Big|^2=frac{K}{2}+frac{1}{4}(|z_1|^2+|z_2|^2-z_1bar{z}_2-bar{z}_1z_2)=frac{K}{2}-frac{1}{4}|z_1-z_2|^2\
Big|z-frac{z_1+z_2}{2}Big|^2=frac{1}{4}Big[2K-|z_1-z_2|^2Big]\
implies 2Kgeq|z_1-z_2|^2implies boxed{Kgeqfrac{1}{2}|z_1-z_2|^2}
$$
As it was asked as a multiple choice question I think my attempt is a bit long and cumbersome. So, is there an easier way to solve this ?
complex-numbers vectors circle
complex-numbers vectors circle
asked Jan 17 at 11:13
ss1729ss1729
1,9331923
asked Jan 17 at 11:13
ss1729ss1729
1,9331923
edited Jan 17 at 19:27
ss1729
asked Jan 17 at 11:13
ss1729ss1729
1,9331923
asked Jan 17 at 11:13
ss1729ss1729
1,9331923
asked Jan 17 at 11:13
ss1729ss1729
1,9331923
1,9331923
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So if point $T$ is for $z$ and $A,B$ for $z_1,z_2$ then we have
$$ AT^2+BT^2=K$$
Let $M = (A+B)/2$ be a midpoint for $AB$ then we have:
$$ K= (T-A)^2+(T-B)^2$$ $$ = 2T^2-2T(A+B)+A^2+B^2 $$ $$ = 2T^2-4Tcdot M + 2M^2- 2M^2+A^2+B^2$$
$$ = 2(T-M)^2+(A-B)^2/2$$
So $$ (T-M)^2= K-(A-B)^2/4$$
and thus $4K geq (A-B)^2$ or c) on your answer sheet.
answered Jan 17 at 11:23
greedoidgreedoid
42k1152105
$endgroup$
1
$begingroup$
Thanx and I just tried to modify OP to better explain my doubt. btw still trying to understand ur proof.
$endgroup$
– ss1729
Jan 17 at 19:40
$begingroup$
You just have to know position vectors.
$endgroup$
– greedoid
Jan 17 at 19:43
$begingroup$
How do u write $(T−A)^2$ for $|z−z1|^2$ ?. are u considering for the case where it is real ?. ould u pls explain a bit more.
$endgroup$
– ss1729
Jan 17 at 20:15
$begingroup$
Do you know vectors?
$endgroup$
– greedoid
Jan 17 at 21:08
$begingroup$
If $T$ is a point with coordinates $(x,y)$ then with same notation $T$ is vector from $(0,0)$ to $(x,y)$. So we have $vec {T} = (x,y) =T$
$endgroup$
– greedoid
Jan 17 at 21:14
|
show 1 more comment
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
So if point $T$ is for $z$ and $A,B$ for $z_1,z_2$ then we have
$$ AT^2+BT^2=K$$
Let $M = (A+B)/2$ be a midpoint for $AB$ then we have:
$$ K= (T-A)^2+(T-B)^2$$ $$ = 2T^2-2T(A+B)+A^2+B^2 $$ $$ = 2T^2-4Tcdot M + 2M^2- 2M^2+A^2+B^2$$
$$ = 2(T-M)^2+(A-B)^2/2$$
So $$ (T-M)^2= K-(A-B)^2/4$$
and thus $4K geq (A-B)^2$ or c) on your answer sheet.
answered Jan 17 at 11:23
greedoidgreedoid
42k1152105
$endgroup$
1
$begingroup$
Thanx and I just tried to modify OP to better explain my doubt. btw still trying to understand ur proof.
$endgroup$
– ss1729
Jan 17 at 19:40
$begingroup$
You just have to know position vectors.
$endgroup$
– greedoid
Jan 17 at 19:43
$begingroup$
How do u write $(T−A)^2$ for $|z−z1|^2$ ?. are u considering for the case where it is real ?. ould u pls explain a bit more.
$endgroup$
– ss1729
Jan 17 at 20:15
$begingroup$
Do you know vectors?
$endgroup$
– greedoid
Jan 17 at 21:08
$begingroup$
If $T$ is a point with coordinates $(x,y)$ then with same notation $T$ is vector from $(0,0)$ to $(x,y)$. So we have $vec {T} = (x,y) =T$
$endgroup$
– greedoid
Jan 17 at 21:14
|
show 1 more comment
$begingroup$
So if point $T$ is for $z$ and $A,B$ for $z_1,z_2$ then we have
$$ AT^2+BT^2=K$$
Let $M = (A+B)/2$ be a midpoint for $AB$ then we have:
$$ K= (T-A)^2+(T-B)^2$$ $$ = 2T^2-2T(A+B)+A^2+B^2 $$ $$ = 2T^2-4Tcdot M + 2M^2- 2M^2+A^2+B^2$$
$$ = 2(T-M)^2+(A-B)^2/2$$
So $$ (T-M)^2= K-(A-B)^2/4$$
and thus $4K geq (A-B)^2$ or c) on your answer sheet.
answered Jan 17 at 11:23
greedoidgreedoid
42k1152105
$endgroup$
1
$begingroup$
Thanx and I just tried to modify OP to better explain my doubt. btw still trying to understand ur proof.
$endgroup$
– ss1729
Jan 17 at 19:40
$begingroup$
You just have to know position vectors.
$endgroup$
– greedoid
Jan 17 at 19:43
$begingroup$
How do u write $(T−A)^2$ for $|z−z1|^2$ ?. are u considering for the case where it is real ?. ould u pls explain a bit more.
$endgroup$
– ss1729
Jan 17 at 20:15
$begingroup$
Do you know vectors?
$endgroup$
– greedoid
Jan 17 at 21:08
$begingroup$
If $T$ is a point with coordinates $(x,y)$ then with same notation $T$ is vector from $(0,0)$ to $(x,y)$. So we have $vec {T} = (x,y) =T$
$endgroup$
– greedoid
Jan 17 at 21:14
|
show 1 more comment
$begingroup$
So if point $T$ is for $z$ and $A,B$ for $z_1,z_2$ then we have
$$ AT^2+BT^2=K$$
Let $M = (A+B)/2$ be a midpoint for $AB$ then we have:
$$ K= (T-A)^2+(T-B)^2$$ $$ = 2T^2-2T(A+B)+A^2+B^2 $$ $$ = 2T^2-4Tcdot M + 2M^2- 2M^2+A^2+B^2$$
$$ = 2(T-M)^2+(A-B)^2/2$$
So $$ (T-M)^2= K-(A-B)^2/4$$
and thus $4K geq (A-B)^2$ or c) on your answer sheet.
answered Jan 17 at 11:23
greedoidgreedoid
42k1152105
$endgroup$
So if point $T$ is for $z$ and $A,B$ for $z_1,z_2$ then we have
$$ AT^2+BT^2=K$$
Let $M = (A+B)/2$ be a midpoint for $AB$ then we have:
$$ K= (T-A)^2+(T-B)^2$$ $$ = 2T^2-2T(A+B)+A^2+B^2 $$ $$ = 2T^2-4Tcdot M + 2M^2- 2M^2+A^2+B^2$$
$$ = 2(T-M)^2+(A-B)^2/2$$
So $$ (T-M)^2= K-(A-B)^2/4$$
and thus $4K geq (A-B)^2$ or c) on your answer sheet.
answered Jan 17 at 11:23
greedoidgreedoid
42k1152105
edited Jan 17 at 19:36
answered Jan 17 at 11:23
greedoidgreedoid
42k1152105
answered Jan 17 at 11:23
greedoidgreedoid
42k1152105
answered Jan 17 at 11:23
greedoidgreedoid
42k1152105
42k1152105
1
$begingroup$
Thanx and I just tried to modify OP to better explain my doubt. btw still trying to understand ur proof.
$endgroup$
– ss1729
Jan 17 at 19:40
$begingroup$
You just have to know position vectors.
$endgroup$
– greedoid
Jan 17 at 19:43
$begingroup$
How do u write $(T−A)^2$ for $|z−z1|^2$ ?. are u considering for the case where it is real ?. ould u pls explain a bit more.
$endgroup$
– ss1729
Jan 17 at 20:15
$begingroup$
Do you know vectors?
$endgroup$
– greedoid
Jan 17 at 21:08
$begingroup$
If $T$ is a point with coordinates $(x,y)$ then with same notation $T$ is vector from $(0,0)$ to $(x,y)$. So we have $vec {T} = (x,y) =T$
$endgroup$
– greedoid
Jan 17 at 21:14
|
show 1 more comment
1
$begingroup$
Thanx and I just tried to modify OP to better explain my doubt. btw still trying to understand ur proof.
$endgroup$
– ss1729
Jan 17 at 19:40
$begingroup$
You just have to know position vectors.
$endgroup$
– greedoid
Jan 17 at 19:43
$begingroup$
How do u write $(T−A)^2$ for $|z−z1|^2$ ?. are u considering for the case where it is real ?. ould u pls explain a bit more.
$endgroup$
– ss1729
Jan 17 at 20:15
$begingroup$
Do you know vectors?
$endgroup$
– greedoid
Jan 17 at 21:08
$begingroup$
If $T$ is a point with coordinates $(x,y)$ then with same notation $T$ is vector from $(0,0)$ to $(x,y)$. So we have $vec {T} = (x,y) =T$
$endgroup$
– greedoid
Jan 17 at 21:14
1
1
$begingroup$
Thanx and I just tried to modify OP to better explain my doubt. btw still trying to understand ur proof.
$endgroup$
– ss1729
Jan 17 at 19:40
$begingroup$
Thanx and I just tried to modify OP to better explain my doubt. btw still trying to understand ur proof.
$endgroup$
– ss1729
Jan 17 at 19:40
$begingroup$
You just have to know position vectors.
$endgroup$
– greedoid
Jan 17 at 19:43
$begingroup$
You just have to know position vectors.
$endgroup$
– greedoid
Jan 17 at 19:43
$begingroup$
How do u write $(T−A)^2$ for $|z−z1|^2$ ?. are u considering for the case where it is real ?. ould u pls explain a bit more.
$endgroup$
– ss1729
Jan 17 at 20:15
$begingroup$
How do u write $(T−A)^2$ for $|z−z1|^2$ ?. are u considering for the case where it is real ?. ould u pls explain a bit more.
$endgroup$
– ss1729
Jan 17 at 20:15
$begingroup$
Do you know vectors?
$endgroup$
– greedoid
Jan 17 at 21:08
$begingroup$
Do you know vectors?
$endgroup$
– greedoid
Jan 17 at 21:08
$begingroup$
If $T$ is a point with coordinates $(x,y)$ then with same notation $T$ is vector from $(0,0)$ to $(x,y)$. So we have $vec {T} = (x,y) =T$
$endgroup$
– greedoid
Jan 17 at 21:14
$begingroup$
If $T$ is a point with coordinates $(x,y)$ then with same notation $T$ is vector from $(0,0)$ to $(x,y)$. So we have $vec {T} = (x,y) =T$
$endgroup$
– greedoid
Jan 17 at 21:14
|
show 1 more comment
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