Prove the locus of intersection of 2 circles on 2 sides of triangle as chords is hyperbola.
$begingroup$
Question
Prove that the locus of the intersection of two equal circles which are described on two sides $EF$ and $EG$ of a triangle as chords is a rectangular hyperbola whose center is the midpoint of $FG$ and which passes through $E$,$F$ and $G$.
Attempt
Consider a triangle $EFG$ with $E(x_1,y_1)$,$F(x_2,y_2)$,$G(x_3,y_3)$ as vertices.
Consider two equal circles,
$$S_1:=x^2+y^2+2g_1x+2f_1y+c_1=0$$
$$S_2:=x^2+y^2+2g_2x+2f_2y+c_2=0$$
Therefore, the condition of equal radii $(r_1=r_2)$ gives,
$$sqrt {g_1^2+f_1^2-c_1^2}=sqrt {g_2^2+f_2^2-c_2^2}$$
As $EF$ is the chord of $S_1$,
$$x_1^2+y_1^2+2g_1x_1+2f_1y_1+c_1=0$$
$$x_2^2+y_2^2+2g_1x_2+2f_1y_2+c_1=0$$
As $EG$ is the chord of $S_2$,
$$x_1^2+y_1^2+2g_2x_1+2f_2y_1+c_2=0$$
$$x_3^2+y_3^2+2g_2x_3+2f_2y_3+c_2=0$$
Let $S_1$ and $S_2$ meet in $(h,k)$. Then we have,
$$h^2+k^2+2g_1h+2f_1k+c_1=0$$
$$h^2+k^2+2g_2h+2f_2k+c_2=0$$
What now I have to do is prove that $(h-frac {x_2+x_3}{2})^2 - (k-frac {y_2+y_3}{2})^2=A^2$ for some constant $A$. One value of $(h,k)$ is clearly $(x_1,y_1)$. Yet neither I can derive the other value of $(h,k)$ nor prove the desired result, for it looks tremendously tedious. Can anyone help me to make the way out?
Edit
Playing with the slope of the common chord of the circles, I find another linear equation in $h$ and $k$,
$$h(f_1-f_2)+k(g_1-g_2)=x_1(f_1-f_2)+y_1(g_1-g_2)$$
As the resulting hyperbola passes through $E$,$F$ and $G$, I have another set of results,
$$(x_1-frac {x_2+x_3}{2})^2 - (y_1-frac {y_2+y_3}{2})^2=A^2$$
$$(x_2-frac {x_2+x_3}{2})^2 - (y_2-frac {y_2+y_3}{2})^2=A^2$$
$$(x_3-frac {x_2+x_3}{2})^2 - (y_3-frac {y_2+y_3}{2})^2=A^2$$
Moreover , as suggested in the comment, I would add that this is the case of a conic through 5 points, $E,F,G,$ the reflections of $E$ in the circumcentre of $∆EFG$ and in the midpoint of $FG$.
analytic-geometry circle conic-sections locus
asked Dec 17 '18 at 5:38
Awe Kumar JhaAwe Kumar Jha
43813
$endgroup$
add a comment |
$begingroup$
Question
Prove that the locus of the intersection of two equal circles which are described on two sides $EF$ and $EG$ of a triangle as chords is a rectangular hyperbola whose center is the midpoint of $FG$ and which passes through $E$,$F$ and $G$.
Attempt
Consider a triangle $EFG$ with $E(x_1,y_1)$,$F(x_2,y_2)$,$G(x_3,y_3)$ as vertices.
Consider two equal circles,
$$S_1:=x^2+y^2+2g_1x+2f_1y+c_1=0$$
$$S_2:=x^2+y^2+2g_2x+2f_2y+c_2=0$$
Therefore, the condition of equal radii $(r_1=r_2)$ gives,
$$sqrt {g_1^2+f_1^2-c_1^2}=sqrt {g_2^2+f_2^2-c_2^2}$$
As $EF$ is the chord of $S_1$,
$$x_1^2+y_1^2+2g_1x_1+2f_1y_1+c_1=0$$
$$x_2^2+y_2^2+2g_1x_2+2f_1y_2+c_1=0$$
As $EG$ is the chord of $S_2$,
$$x_1^2+y_1^2+2g_2x_1+2f_2y_1+c_2=0$$
$$x_3^2+y_3^2+2g_2x_3+2f_2y_3+c_2=0$$
Let $S_1$ and $S_2$ meet in $(h,k)$. Then we have,
$$h^2+k^2+2g_1h+2f_1k+c_1=0$$
$$h^2+k^2+2g_2h+2f_2k+c_2=0$$
What now I have to do is prove that $(h-frac {x_2+x_3}{2})^2 - (k-frac {y_2+y_3}{2})^2=A^2$ for some constant $A$. One value of $(h,k)$ is clearly $(x_1,y_1)$. Yet neither I can derive the other value of $(h,k)$ nor prove the desired result, for it looks tremendously tedious. Can anyone help me to make the way out?
Edit
Playing with the slope of the common chord of the circles, I find another linear equation in $h$ and $k$,
$$h(f_1-f_2)+k(g_1-g_2)=x_1(f_1-f_2)+y_1(g_1-g_2)$$
As the resulting hyperbola passes through $E$,$F$ and $G$, I have another set of results,
$$(x_1-frac {x_2+x_3}{2})^2 - (y_1-frac {y_2+y_3}{2})^2=A^2$$
$$(x_2-frac {x_2+x_3}{2})^2 - (y_2-frac {y_2+y_3}{2})^2=A^2$$
$$(x_3-frac {x_2+x_3}{2})^2 - (y_3-frac {y_2+y_3}{2})^2=A^2$$
Moreover , as suggested in the comment, I would add that this is the case of a conic through 5 points, $E,F,G,$ the reflections of $E$ in the circumcentre of $∆EFG$ and in the midpoint of $FG$.
analytic-geometry circle conic-sections locus
asked Dec 17 '18 at 5:38
Awe Kumar JhaAwe Kumar Jha
43813
$endgroup$
1
$begingroup$
I believe this is the $5$-point conic through $E$, $F$, $G$, the reflection of $E$ in the circumcenter of $triangle EFG$, and the reflection of $E$ in the midpoint of $overline{FG}$. The center is the midpoint of $overline{FG}$. I got this through some ugly coordinate-bashing in Mathematica. I'm seeking a cleaner derivation, along with a showing that the resulting hyperbola is rectangular.
$endgroup$
– Blue
Dec 17 '18 at 8:55
$begingroup$
@Blue, have you tried parametric forms , I suspect if they can be of a use.
$endgroup$
– Awe Kumar Jha
Dec 26 '18 at 8:38
add a comment |
$begingroup$
Question
Prove that the locus of the intersection of two equal circles which are described on two sides $EF$ and $EG$ of a triangle as chords is a rectangular hyperbola whose center is the midpoint of $FG$ and which passes through $E$,$F$ and $G$.
Attempt
Consider a triangle $EFG$ with $E(x_1,y_1)$,$F(x_2,y_2)$,$G(x_3,y_3)$ as vertices.
Consider two equal circles,
$$S_1:=x^2+y^2+2g_1x+2f_1y+c_1=0$$
$$S_2:=x^2+y^2+2g_2x+2f_2y+c_2=0$$
Therefore, the condition of equal radii $(r_1=r_2)$ gives,
$$sqrt {g_1^2+f_1^2-c_1^2}=sqrt {g_2^2+f_2^2-c_2^2}$$
As $EF$ is the chord of $S_1$,
$$x_1^2+y_1^2+2g_1x_1+2f_1y_1+c_1=0$$
$$x_2^2+y_2^2+2g_1x_2+2f_1y_2+c_1=0$$
As $EG$ is the chord of $S_2$,
$$x_1^2+y_1^2+2g_2x_1+2f_2y_1+c_2=0$$
$$x_3^2+y_3^2+2g_2x_3+2f_2y_3+c_2=0$$
Let $S_1$ and $S_2$ meet in $(h,k)$. Then we have,
$$h^2+k^2+2g_1h+2f_1k+c_1=0$$
$$h^2+k^2+2g_2h+2f_2k+c_2=0$$
What now I have to do is prove that $(h-frac {x_2+x_3}{2})^2 - (k-frac {y_2+y_3}{2})^2=A^2$ for some constant $A$. One value of $(h,k)$ is clearly $(x_1,y_1)$. Yet neither I can derive the other value of $(h,k)$ nor prove the desired result, for it looks tremendously tedious. Can anyone help me to make the way out?
Edit
Playing with the slope of the common chord of the circles, I find another linear equation in $h$ and $k$,
$$h(f_1-f_2)+k(g_1-g_2)=x_1(f_1-f_2)+y_1(g_1-g_2)$$
As the resulting hyperbola passes through $E$,$F$ and $G$, I have another set of results,
$$(x_1-frac {x_2+x_3}{2})^2 - (y_1-frac {y_2+y_3}{2})^2=A^2$$
$$(x_2-frac {x_2+x_3}{2})^2 - (y_2-frac {y_2+y_3}{2})^2=A^2$$
$$(x_3-frac {x_2+x_3}{2})^2 - (y_3-frac {y_2+y_3}{2})^2=A^2$$
Moreover , as suggested in the comment, I would add that this is the case of a conic through 5 points, $E,F,G,$ the reflections of $E$ in the circumcentre of $∆EFG$ and in the midpoint of $FG$.
analytic-geometry circle conic-sections locus
asked Dec 17 '18 at 5:38
Awe Kumar JhaAwe Kumar Jha
43813
$endgroup$
Question
Prove that the locus of the intersection of two equal circles which are described on two sides $EF$ and $EG$ of a triangle as chords is a rectangular hyperbola whose center is the midpoint of $FG$ and which passes through $E$,$F$ and $G$.
Attempt
Consider a triangle $EFG$ with $E(x_1,y_1)$,$F(x_2,y_2)$,$G(x_3,y_3)$ as vertices.
Consider two equal circles,
$$S_1:=x^2+y^2+2g_1x+2f_1y+c_1=0$$
$$S_2:=x^2+y^2+2g_2x+2f_2y+c_2=0$$
Therefore, the condition of equal radii $(r_1=r_2)$ gives,
$$sqrt {g_1^2+f_1^2-c_1^2}=sqrt {g_2^2+f_2^2-c_2^2}$$
As $EF$ is the chord of $S_1$,
$$x_1^2+y_1^2+2g_1x_1+2f_1y_1+c_1=0$$
$$x_2^2+y_2^2+2g_1x_2+2f_1y_2+c_1=0$$
As $EG$ is the chord of $S_2$,
$$x_1^2+y_1^2+2g_2x_1+2f_2y_1+c_2=0$$
$$x_3^2+y_3^2+2g_2x_3+2f_2y_3+c_2=0$$
Let $S_1$ and $S_2$ meet in $(h,k)$. Then we have,
$$h^2+k^2+2g_1h+2f_1k+c_1=0$$
$$h^2+k^2+2g_2h+2f_2k+c_2=0$$
What now I have to do is prove that $(h-frac {x_2+x_3}{2})^2 - (k-frac {y_2+y_3}{2})^2=A^2$ for some constant $A$. One value of $(h,k)$ is clearly $(x_1,y_1)$. Yet neither I can derive the other value of $(h,k)$ nor prove the desired result, for it looks tremendously tedious. Can anyone help me to make the way out?
Edit
Playing with the slope of the common chord of the circles, I find another linear equation in $h$ and $k$,
$$h(f_1-f_2)+k(g_1-g_2)=x_1(f_1-f_2)+y_1(g_1-g_2)$$
As the resulting hyperbola passes through $E$,$F$ and $G$, I have another set of results,
$$(x_1-frac {x_2+x_3}{2})^2 - (y_1-frac {y_2+y_3}{2})^2=A^2$$
$$(x_2-frac {x_2+x_3}{2})^2 - (y_2-frac {y_2+y_3}{2})^2=A^2$$
$$(x_3-frac {x_2+x_3}{2})^2 - (y_3-frac {y_2+y_3}{2})^2=A^2$$
Moreover , as suggested in the comment, I would add that this is the case of a conic through 5 points, $E,F,G,$ the reflections of $E$ in the circumcentre of $∆EFG$ and in the midpoint of $FG$.
analytic-geometry circle conic-sections locus
analytic-geometry circle conic-sections locus
asked Dec 17 '18 at 5:38
Awe Kumar JhaAwe Kumar Jha
43813
asked Dec 17 '18 at 5:38
Awe Kumar JhaAwe Kumar Jha
43813
edited Dec 24 '18 at 7:47
Awe Kumar Jha
asked Dec 17 '18 at 5:38
Awe Kumar JhaAwe Kumar Jha
43813
asked Dec 17 '18 at 5:38
Awe Kumar JhaAwe Kumar Jha
43813
asked Dec 17 '18 at 5:38
Awe Kumar JhaAwe Kumar Jha
43813
43813
1
$begingroup$
I believe this is the $5$-point conic through $E$, $F$, $G$, the reflection of $E$ in the circumcenter of $triangle EFG$, and the reflection of $E$ in the midpoint of $overline{FG}$. The center is the midpoint of $overline{FG}$. I got this through some ugly coordinate-bashing in Mathematica. I'm seeking a cleaner derivation, along with a showing that the resulting hyperbola is rectangular.
$endgroup$
– Blue
Dec 17 '18 at 8:55
$begingroup$
@Blue, have you tried parametric forms , I suspect if they can be of a use.
$endgroup$
– Awe Kumar Jha
Dec 26 '18 at 8:38
add a comment |
1
$begingroup$
I believe this is the $5$-point conic through $E$, $F$, $G$, the reflection of $E$ in the circumcenter of $triangle EFG$, and the reflection of $E$ in the midpoint of $overline{FG}$. The center is the midpoint of $overline{FG}$. I got this through some ugly coordinate-bashing in Mathematica. I'm seeking a cleaner derivation, along with a showing that the resulting hyperbola is rectangular.
$endgroup$
– Blue
Dec 17 '18 at 8:55
$begingroup$
@Blue, have you tried parametric forms , I suspect if they can be of a use.
$endgroup$
– Awe Kumar Jha
Dec 26 '18 at 8:38
1
1
$begingroup$
I believe this is the $5$-point conic through $E$, $F$, $G$, the reflection of $E$ in the circumcenter of $triangle EFG$, and the reflection of $E$ in the midpoint of $overline{FG}$. The center is the midpoint of $overline{FG}$. I got this through some ugly coordinate-bashing in Mathematica. I'm seeking a cleaner derivation, along with a showing that the resulting hyperbola is rectangular.
$endgroup$
– Blue
Dec 17 '18 at 8:55
$begingroup$
I believe this is the $5$-point conic through $E$, $F$, $G$, the reflection of $E$ in the circumcenter of $triangle EFG$, and the reflection of $E$ in the midpoint of $overline{FG}$. The center is the midpoint of $overline{FG}$. I got this through some ugly coordinate-bashing in Mathematica. I'm seeking a cleaner derivation, along with a showing that the resulting hyperbola is rectangular.
$endgroup$
– Blue
Dec 17 '18 at 8:55
$begingroup$
@Blue, have you tried parametric forms , I suspect if they can be of a use.
$endgroup$
– Awe Kumar Jha
Dec 26 '18 at 8:38
$begingroup$
@Blue, have you tried parametric forms , I suspect if they can be of a use.
$endgroup$
– Awe Kumar Jha
Dec 26 '18 at 8:38
add a comment |
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I believe this is the $5$-point conic through $E$, $F$, $G$, the reflection of $E$ in the circumcenter of $triangle EFG$, and the reflection of $E$ in the midpoint of $overline{FG}$. The center is the midpoint of $overline{FG}$. I got this through some ugly coordinate-bashing in Mathematica. I'm seeking a cleaner derivation, along with a showing that the resulting hyperbola is rectangular.
$endgroup$
– Blue
Dec 17 '18 at 8:55
$begingroup$
@Blue, have you tried parametric forms , I suspect if they can be of a use.
$endgroup$
– Awe Kumar Jha
Dec 26 '18 at 8:38