Magnitude of Complex Number And Other Equation
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Sorry for the unclear question, but I have no idea how to solve this problem:
For certain positive integers $a,b,c$ and $d$, the set of complex numbers $z$ that satisfy $|z-5sqrt{3}-5i|=5$ is equivalent to the set of complex numbers $z$ that satisfy $left| frac{1}{z} - frac{1}{a sqrt{b}} + frac{i}{c} right| = frac{1}{d}$, where $b$ is not divisible by the square of a prime. Find $a+b+c+d$.
My approach was to graph a circle on the imaginary plane with center $(5sqrt{3},5)$ with radius 5. The equation of this circle is $(x-5sqrt{3})^2+(y-5i)^2=25$ (I think?). I'm not sure how to relate the set of $z$ to the equation. Any help is appreciated.
algebra-precalculus complex-numbers
$endgroup$
add a comment |
$begingroup$
Sorry for the unclear question, but I have no idea how to solve this problem:
For certain positive integers $a,b,c$ and $d$, the set of complex numbers $z$ that satisfy $|z-5sqrt{3}-5i|=5$ is equivalent to the set of complex numbers $z$ that satisfy $left| frac{1}{z} - frac{1}{a sqrt{b}} + frac{i}{c} right| = frac{1}{d}$, where $b$ is not divisible by the square of a prime. Find $a+b+c+d$.
My approach was to graph a circle on the imaginary plane with center $(5sqrt{3},5)$ with radius 5. The equation of this circle is $(x-5sqrt{3})^2+(y-5i)^2=25$ (I think?). I'm not sure how to relate the set of $z$ to the equation. Any help is appreciated.
algebra-precalculus complex-numbers
$endgroup$
add a comment |
$begingroup$
Sorry for the unclear question, but I have no idea how to solve this problem:
For certain positive integers $a,b,c$ and $d$, the set of complex numbers $z$ that satisfy $|z-5sqrt{3}-5i|=5$ is equivalent to the set of complex numbers $z$ that satisfy $left| frac{1}{z} - frac{1}{a sqrt{b}} + frac{i}{c} right| = frac{1}{d}$, where $b$ is not divisible by the square of a prime. Find $a+b+c+d$.
My approach was to graph a circle on the imaginary plane with center $(5sqrt{3},5)$ with radius 5. The equation of this circle is $(x-5sqrt{3})^2+(y-5i)^2=25$ (I think?). I'm not sure how to relate the set of $z$ to the equation. Any help is appreciated.
algebra-precalculus complex-numbers
$endgroup$
Sorry for the unclear question, but I have no idea how to solve this problem:
For certain positive integers $a,b,c$ and $d$, the set of complex numbers $z$ that satisfy $|z-5sqrt{3}-5i|=5$ is equivalent to the set of complex numbers $z$ that satisfy $left| frac{1}{z} - frac{1}{a sqrt{b}} + frac{i}{c} right| = frac{1}{d}$, where $b$ is not divisible by the square of a prime. Find $a+b+c+d$.
My approach was to graph a circle on the imaginary plane with center $(5sqrt{3},5)$ with radius 5. The equation of this circle is $(x-5sqrt{3})^2+(y-5i)^2=25$ (I think?). I'm not sure how to relate the set of $z$ to the equation. Any help is appreciated.
algebra-precalculus complex-numbers
algebra-precalculus complex-numbers
asked Jan 1 at 23:47
A Piercing ArrowA Piercing Arrow
538117
538117
add a comment |
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1 Answer
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$begingroup$
Note that if you replace $frac 1z$ with $u$ in your second equation $left| frac{1}{z} - frac{1}{a sqrt{b}} + frac{i}{c} right| = frac{1}{d}$ then the locus for $u$ in the complex plane is another circle whose radius is $frac 1d$ and center is $frac{1}{z} - frac{1}{a sqrt{b}}$. Thus the desired circle is the "inversion" of your first circle under the operation $z to frac 1z$.
You have the equation of the given circle, so find the equation of the inverted circle. There are several ways to do this. One way is to find three points (complex numbers) on the original circle, find the reciprocals of those three points, then find the circle that goes through those three points. Once you find that circle you find its center and radius then can find the values of $a, b, c,$ and $d$.
Or better, find the point on the first circle that is closest to the origin (point $0+0i$) and the point that is farthest from the origin. Those two points are the endpoints of a diameter of the first circle. The closest point is $frac 52sqrt{3} + frac 52i$ and the farthest is $frac {15}2sqrt{3} + frac {15}2i$, which can be seen easily geometrically (see the diagram, and recognize that the right triangle is a 30°-60°-90° triangle). Now invert those two points. The closest point on the first circle goes to the farthest point in the inverted circle, and vice versa. So the inverted points determine a diameter of the inverted circle. The average of those two inverted points is then the diameter of the inverted circle, and the radius is the modulus of the difference between the center and either of the two inverted points. This is much easier than using three general points on the circles.
Here are the final answers: when we invert the near point $z_1=frac 52sqrt{3} + frac 52i$ we get $u_1=frac 1{10}sqrt 3-frac 1{10}i$ and when we invert the far point $z_2=frac {15}2sqrt{3} + frac {15}2i$ we get $u_2=frac 1{30}sqrt 3-frac 1{30}i$. The center of the inverted circle is the average of those two points, $u_c=frac 1{15}sqrt 3-frac 1{15}i$. If we rationalize the numerator we get
$$u_c=frac 1{5sqrt 3}-frac 1{15}i$$
The radius of the circle is
$$|u_1-u_c|=sqrt{left[frac 1{10}-frac 1{15}right]^2sqrt 3^2+left[frac 1{10}-frac 1{15}right]^2}=sqrt{frac 3{900}+frac 1{900}}=frac 1{15}$$
So the final result is
$$a+b+c+d = 5+3+15+15 = 38$$
$endgroup$
$begingroup$
Let's say you have a point of $p+qi$. How would we "invert" this to follow the given equation. I understand the help, but I'm still not sure how to invert.
$endgroup$
– A Piercing Arrow
Jan 2 at 1:25
$begingroup$
@APiercingArrow: If the point $p+qi$ is on the first circle, then the point $frac 1{p+qi}$ is on the inverted circle. Find the real and imaginary components of $frac 1{p+qi}$ in the usual way: multiply the top and the bottom of the fraction by the conjugate of the denominator.
$endgroup$
– Rory Daulton
Jan 2 at 1:28
$begingroup$
I used the points $5sqrt{3}/2 + frac{5i}{2}$ (which inverted to $sqrt{3}/10 - I/10$), $5sqrt{3} + 0i$ (which inverted to $sqrt{3}/15 + 0i$), and $5sqrt{3}+10i$ (which inverted to $frac{2sqrt{3}-8i}{95}$. Are these correct points, and should I continue to solve for the circle?
$endgroup$
– A Piercing Arrow
Jan 2 at 19:19
$begingroup$
@APiercingArrow: Those three points are correct and your first two inversions are correct, but the third inversion is wrong. But ignore all but the first point and its inversion--I edited my answer to show a way using only two points to get your answer. One of those two points is your first point, and I state the second point. Continue from there and ask if you still need help.
$endgroup$
– Rory Daulton
Jan 2 at 23:16
$begingroup$
Thank you so much! I got the equation of the circle to be $(x -sqrt{3}/15$ )^2 + (y+i/15)^2 = 13/(225^2)$. With this equation, $a = 5$, $b=3$, $c=15$, and $d$ is not an integer. Where did I go wrong?
$endgroup$
– A Piercing Arrow
Jan 3 at 22:32
|
show 1 more comment
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1 Answer
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$begingroup$
Note that if you replace $frac 1z$ with $u$ in your second equation $left| frac{1}{z} - frac{1}{a sqrt{b}} + frac{i}{c} right| = frac{1}{d}$ then the locus for $u$ in the complex plane is another circle whose radius is $frac 1d$ and center is $frac{1}{z} - frac{1}{a sqrt{b}}$. Thus the desired circle is the "inversion" of your first circle under the operation $z to frac 1z$.
You have the equation of the given circle, so find the equation of the inverted circle. There are several ways to do this. One way is to find three points (complex numbers) on the original circle, find the reciprocals of those three points, then find the circle that goes through those three points. Once you find that circle you find its center and radius then can find the values of $a, b, c,$ and $d$.
Or better, find the point on the first circle that is closest to the origin (point $0+0i$) and the point that is farthest from the origin. Those two points are the endpoints of a diameter of the first circle. The closest point is $frac 52sqrt{3} + frac 52i$ and the farthest is $frac {15}2sqrt{3} + frac {15}2i$, which can be seen easily geometrically (see the diagram, and recognize that the right triangle is a 30°-60°-90° triangle). Now invert those two points. The closest point on the first circle goes to the farthest point in the inverted circle, and vice versa. So the inverted points determine a diameter of the inverted circle. The average of those two inverted points is then the diameter of the inverted circle, and the radius is the modulus of the difference between the center and either of the two inverted points. This is much easier than using three general points on the circles.
Here are the final answers: when we invert the near point $z_1=frac 52sqrt{3} + frac 52i$ we get $u_1=frac 1{10}sqrt 3-frac 1{10}i$ and when we invert the far point $z_2=frac {15}2sqrt{3} + frac {15}2i$ we get $u_2=frac 1{30}sqrt 3-frac 1{30}i$. The center of the inverted circle is the average of those two points, $u_c=frac 1{15}sqrt 3-frac 1{15}i$. If we rationalize the numerator we get
$$u_c=frac 1{5sqrt 3}-frac 1{15}i$$
The radius of the circle is
$$|u_1-u_c|=sqrt{left[frac 1{10}-frac 1{15}right]^2sqrt 3^2+left[frac 1{10}-frac 1{15}right]^2}=sqrt{frac 3{900}+frac 1{900}}=frac 1{15}$$
So the final result is
$$a+b+c+d = 5+3+15+15 = 38$$
$endgroup$
$begingroup$
Let's say you have a point of $p+qi$. How would we "invert" this to follow the given equation. I understand the help, but I'm still not sure how to invert.
$endgroup$
– A Piercing Arrow
Jan 2 at 1:25
$begingroup$
@APiercingArrow: If the point $p+qi$ is on the first circle, then the point $frac 1{p+qi}$ is on the inverted circle. Find the real and imaginary components of $frac 1{p+qi}$ in the usual way: multiply the top and the bottom of the fraction by the conjugate of the denominator.
$endgroup$
– Rory Daulton
Jan 2 at 1:28
$begingroup$
I used the points $5sqrt{3}/2 + frac{5i}{2}$ (which inverted to $sqrt{3}/10 - I/10$), $5sqrt{3} + 0i$ (which inverted to $sqrt{3}/15 + 0i$), and $5sqrt{3}+10i$ (which inverted to $frac{2sqrt{3}-8i}{95}$. Are these correct points, and should I continue to solve for the circle?
$endgroup$
– A Piercing Arrow
Jan 2 at 19:19
$begingroup$
@APiercingArrow: Those three points are correct and your first two inversions are correct, but the third inversion is wrong. But ignore all but the first point and its inversion--I edited my answer to show a way using only two points to get your answer. One of those two points is your first point, and I state the second point. Continue from there and ask if you still need help.
$endgroup$
– Rory Daulton
Jan 2 at 23:16
$begingroup$
Thank you so much! I got the equation of the circle to be $(x -sqrt{3}/15$ )^2 + (y+i/15)^2 = 13/(225^2)$. With this equation, $a = 5$, $b=3$, $c=15$, and $d$ is not an integer. Where did I go wrong?
$endgroup$
– A Piercing Arrow
Jan 3 at 22:32
|
show 1 more comment
$begingroup$
Note that if you replace $frac 1z$ with $u$ in your second equation $left| frac{1}{z} - frac{1}{a sqrt{b}} + frac{i}{c} right| = frac{1}{d}$ then the locus for $u$ in the complex plane is another circle whose radius is $frac 1d$ and center is $frac{1}{z} - frac{1}{a sqrt{b}}$. Thus the desired circle is the "inversion" of your first circle under the operation $z to frac 1z$.
You have the equation of the given circle, so find the equation of the inverted circle. There are several ways to do this. One way is to find three points (complex numbers) on the original circle, find the reciprocals of those three points, then find the circle that goes through those three points. Once you find that circle you find its center and radius then can find the values of $a, b, c,$ and $d$.
Or better, find the point on the first circle that is closest to the origin (point $0+0i$) and the point that is farthest from the origin. Those two points are the endpoints of a diameter of the first circle. The closest point is $frac 52sqrt{3} + frac 52i$ and the farthest is $frac {15}2sqrt{3} + frac {15}2i$, which can be seen easily geometrically (see the diagram, and recognize that the right triangle is a 30°-60°-90° triangle). Now invert those two points. The closest point on the first circle goes to the farthest point in the inverted circle, and vice versa. So the inverted points determine a diameter of the inverted circle. The average of those two inverted points is then the diameter of the inverted circle, and the radius is the modulus of the difference between the center and either of the two inverted points. This is much easier than using three general points on the circles.
Here are the final answers: when we invert the near point $z_1=frac 52sqrt{3} + frac 52i$ we get $u_1=frac 1{10}sqrt 3-frac 1{10}i$ and when we invert the far point $z_2=frac {15}2sqrt{3} + frac {15}2i$ we get $u_2=frac 1{30}sqrt 3-frac 1{30}i$. The center of the inverted circle is the average of those two points, $u_c=frac 1{15}sqrt 3-frac 1{15}i$. If we rationalize the numerator we get
$$u_c=frac 1{5sqrt 3}-frac 1{15}i$$
The radius of the circle is
$$|u_1-u_c|=sqrt{left[frac 1{10}-frac 1{15}right]^2sqrt 3^2+left[frac 1{10}-frac 1{15}right]^2}=sqrt{frac 3{900}+frac 1{900}}=frac 1{15}$$
So the final result is
$$a+b+c+d = 5+3+15+15 = 38$$
$endgroup$
$begingroup$
Let's say you have a point of $p+qi$. How would we "invert" this to follow the given equation. I understand the help, but I'm still not sure how to invert.
$endgroup$
– A Piercing Arrow
Jan 2 at 1:25
$begingroup$
@APiercingArrow: If the point $p+qi$ is on the first circle, then the point $frac 1{p+qi}$ is on the inverted circle. Find the real and imaginary components of $frac 1{p+qi}$ in the usual way: multiply the top and the bottom of the fraction by the conjugate of the denominator.
$endgroup$
– Rory Daulton
Jan 2 at 1:28
$begingroup$
I used the points $5sqrt{3}/2 + frac{5i}{2}$ (which inverted to $sqrt{3}/10 - I/10$), $5sqrt{3} + 0i$ (which inverted to $sqrt{3}/15 + 0i$), and $5sqrt{3}+10i$ (which inverted to $frac{2sqrt{3}-8i}{95}$. Are these correct points, and should I continue to solve for the circle?
$endgroup$
– A Piercing Arrow
Jan 2 at 19:19
$begingroup$
@APiercingArrow: Those three points are correct and your first two inversions are correct, but the third inversion is wrong. But ignore all but the first point and its inversion--I edited my answer to show a way using only two points to get your answer. One of those two points is your first point, and I state the second point. Continue from there and ask if you still need help.
$endgroup$
– Rory Daulton
Jan 2 at 23:16
$begingroup$
Thank you so much! I got the equation of the circle to be $(x -sqrt{3}/15$ )^2 + (y+i/15)^2 = 13/(225^2)$. With this equation, $a = 5$, $b=3$, $c=15$, and $d$ is not an integer. Where did I go wrong?
$endgroup$
– A Piercing Arrow
Jan 3 at 22:32
|
show 1 more comment
$begingroup$
Note that if you replace $frac 1z$ with $u$ in your second equation $left| frac{1}{z} - frac{1}{a sqrt{b}} + frac{i}{c} right| = frac{1}{d}$ then the locus for $u$ in the complex plane is another circle whose radius is $frac 1d$ and center is $frac{1}{z} - frac{1}{a sqrt{b}}$. Thus the desired circle is the "inversion" of your first circle under the operation $z to frac 1z$.
You have the equation of the given circle, so find the equation of the inverted circle. There are several ways to do this. One way is to find three points (complex numbers) on the original circle, find the reciprocals of those three points, then find the circle that goes through those three points. Once you find that circle you find its center and radius then can find the values of $a, b, c,$ and $d$.
Or better, find the point on the first circle that is closest to the origin (point $0+0i$) and the point that is farthest from the origin. Those two points are the endpoints of a diameter of the first circle. The closest point is $frac 52sqrt{3} + frac 52i$ and the farthest is $frac {15}2sqrt{3} + frac {15}2i$, which can be seen easily geometrically (see the diagram, and recognize that the right triangle is a 30°-60°-90° triangle). Now invert those two points. The closest point on the first circle goes to the farthest point in the inverted circle, and vice versa. So the inverted points determine a diameter of the inverted circle. The average of those two inverted points is then the diameter of the inverted circle, and the radius is the modulus of the difference between the center and either of the two inverted points. This is much easier than using three general points on the circles.
Here are the final answers: when we invert the near point $z_1=frac 52sqrt{3} + frac 52i$ we get $u_1=frac 1{10}sqrt 3-frac 1{10}i$ and when we invert the far point $z_2=frac {15}2sqrt{3} + frac {15}2i$ we get $u_2=frac 1{30}sqrt 3-frac 1{30}i$. The center of the inverted circle is the average of those two points, $u_c=frac 1{15}sqrt 3-frac 1{15}i$. If we rationalize the numerator we get
$$u_c=frac 1{5sqrt 3}-frac 1{15}i$$
The radius of the circle is
$$|u_1-u_c|=sqrt{left[frac 1{10}-frac 1{15}right]^2sqrt 3^2+left[frac 1{10}-frac 1{15}right]^2}=sqrt{frac 3{900}+frac 1{900}}=frac 1{15}$$
So the final result is
$$a+b+c+d = 5+3+15+15 = 38$$
$endgroup$
Note that if you replace $frac 1z$ with $u$ in your second equation $left| frac{1}{z} - frac{1}{a sqrt{b}} + frac{i}{c} right| = frac{1}{d}$ then the locus for $u$ in the complex plane is another circle whose radius is $frac 1d$ and center is $frac{1}{z} - frac{1}{a sqrt{b}}$. Thus the desired circle is the "inversion" of your first circle under the operation $z to frac 1z$.
You have the equation of the given circle, so find the equation of the inverted circle. There are several ways to do this. One way is to find three points (complex numbers) on the original circle, find the reciprocals of those three points, then find the circle that goes through those three points. Once you find that circle you find its center and radius then can find the values of $a, b, c,$ and $d$.
Or better, find the point on the first circle that is closest to the origin (point $0+0i$) and the point that is farthest from the origin. Those two points are the endpoints of a diameter of the first circle. The closest point is $frac 52sqrt{3} + frac 52i$ and the farthest is $frac {15}2sqrt{3} + frac {15}2i$, which can be seen easily geometrically (see the diagram, and recognize that the right triangle is a 30°-60°-90° triangle). Now invert those two points. The closest point on the first circle goes to the farthest point in the inverted circle, and vice versa. So the inverted points determine a diameter of the inverted circle. The average of those two inverted points is then the diameter of the inverted circle, and the radius is the modulus of the difference between the center and either of the two inverted points. This is much easier than using three general points on the circles.
Here are the final answers: when we invert the near point $z_1=frac 52sqrt{3} + frac 52i$ we get $u_1=frac 1{10}sqrt 3-frac 1{10}i$ and when we invert the far point $z_2=frac {15}2sqrt{3} + frac {15}2i$ we get $u_2=frac 1{30}sqrt 3-frac 1{30}i$. The center of the inverted circle is the average of those two points, $u_c=frac 1{15}sqrt 3-frac 1{15}i$. If we rationalize the numerator we get
$$u_c=frac 1{5sqrt 3}-frac 1{15}i$$
The radius of the circle is
$$|u_1-u_c|=sqrt{left[frac 1{10}-frac 1{15}right]^2sqrt 3^2+left[frac 1{10}-frac 1{15}right]^2}=sqrt{frac 3{900}+frac 1{900}}=frac 1{15}$$
So the final result is
$$a+b+c+d = 5+3+15+15 = 38$$
edited Jan 5 at 18:04
answered Jan 2 at 1:08
Rory DaultonRory Daulton
29.4k53254
29.4k53254
$begingroup$
Let's say you have a point of $p+qi$. How would we "invert" this to follow the given equation. I understand the help, but I'm still not sure how to invert.
$endgroup$
– A Piercing Arrow
Jan 2 at 1:25
$begingroup$
@APiercingArrow: If the point $p+qi$ is on the first circle, then the point $frac 1{p+qi}$ is on the inverted circle. Find the real and imaginary components of $frac 1{p+qi}$ in the usual way: multiply the top and the bottom of the fraction by the conjugate of the denominator.
$endgroup$
– Rory Daulton
Jan 2 at 1:28
$begingroup$
I used the points $5sqrt{3}/2 + frac{5i}{2}$ (which inverted to $sqrt{3}/10 - I/10$), $5sqrt{3} + 0i$ (which inverted to $sqrt{3}/15 + 0i$), and $5sqrt{3}+10i$ (which inverted to $frac{2sqrt{3}-8i}{95}$. Are these correct points, and should I continue to solve for the circle?
$endgroup$
– A Piercing Arrow
Jan 2 at 19:19
$begingroup$
@APiercingArrow: Those three points are correct and your first two inversions are correct, but the third inversion is wrong. But ignore all but the first point and its inversion--I edited my answer to show a way using only two points to get your answer. One of those two points is your first point, and I state the second point. Continue from there and ask if you still need help.
$endgroup$
– Rory Daulton
Jan 2 at 23:16
$begingroup$
Thank you so much! I got the equation of the circle to be $(x -sqrt{3}/15$ )^2 + (y+i/15)^2 = 13/(225^2)$. With this equation, $a = 5$, $b=3$, $c=15$, and $d$ is not an integer. Where did I go wrong?
$endgroup$
– A Piercing Arrow
Jan 3 at 22:32
|
show 1 more comment
$begingroup$
Let's say you have a point of $p+qi$. How would we "invert" this to follow the given equation. I understand the help, but I'm still not sure how to invert.
$endgroup$
– A Piercing Arrow
Jan 2 at 1:25
$begingroup$
@APiercingArrow: If the point $p+qi$ is on the first circle, then the point $frac 1{p+qi}$ is on the inverted circle. Find the real and imaginary components of $frac 1{p+qi}$ in the usual way: multiply the top and the bottom of the fraction by the conjugate of the denominator.
$endgroup$
– Rory Daulton
Jan 2 at 1:28
$begingroup$
I used the points $5sqrt{3}/2 + frac{5i}{2}$ (which inverted to $sqrt{3}/10 - I/10$), $5sqrt{3} + 0i$ (which inverted to $sqrt{3}/15 + 0i$), and $5sqrt{3}+10i$ (which inverted to $frac{2sqrt{3}-8i}{95}$. Are these correct points, and should I continue to solve for the circle?
$endgroup$
– A Piercing Arrow
Jan 2 at 19:19
$begingroup$
@APiercingArrow: Those three points are correct and your first two inversions are correct, but the third inversion is wrong. But ignore all but the first point and its inversion--I edited my answer to show a way using only two points to get your answer. One of those two points is your first point, and I state the second point. Continue from there and ask if you still need help.
$endgroup$
– Rory Daulton
Jan 2 at 23:16
$begingroup$
Thank you so much! I got the equation of the circle to be $(x -sqrt{3}/15$ )^2 + (y+i/15)^2 = 13/(225^2)$. With this equation, $a = 5$, $b=3$, $c=15$, and $d$ is not an integer. Where did I go wrong?
$endgroup$
– A Piercing Arrow
Jan 3 at 22:32
$begingroup$
Let's say you have a point of $p+qi$. How would we "invert" this to follow the given equation. I understand the help, but I'm still not sure how to invert.
$endgroup$
– A Piercing Arrow
Jan 2 at 1:25
$begingroup$
Let's say you have a point of $p+qi$. How would we "invert" this to follow the given equation. I understand the help, but I'm still not sure how to invert.
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– A Piercing Arrow
Jan 2 at 1:25
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@APiercingArrow: If the point $p+qi$ is on the first circle, then the point $frac 1{p+qi}$ is on the inverted circle. Find the real and imaginary components of $frac 1{p+qi}$ in the usual way: multiply the top and the bottom of the fraction by the conjugate of the denominator.
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– Rory Daulton
Jan 2 at 1:28
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@APiercingArrow: If the point $p+qi$ is on the first circle, then the point $frac 1{p+qi}$ is on the inverted circle. Find the real and imaginary components of $frac 1{p+qi}$ in the usual way: multiply the top and the bottom of the fraction by the conjugate of the denominator.
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– Rory Daulton
Jan 2 at 1:28
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I used the points $5sqrt{3}/2 + frac{5i}{2}$ (which inverted to $sqrt{3}/10 - I/10$), $5sqrt{3} + 0i$ (which inverted to $sqrt{3}/15 + 0i$), and $5sqrt{3}+10i$ (which inverted to $frac{2sqrt{3}-8i}{95}$. Are these correct points, and should I continue to solve for the circle?
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– A Piercing Arrow
Jan 2 at 19:19
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I used the points $5sqrt{3}/2 + frac{5i}{2}$ (which inverted to $sqrt{3}/10 - I/10$), $5sqrt{3} + 0i$ (which inverted to $sqrt{3}/15 + 0i$), and $5sqrt{3}+10i$ (which inverted to $frac{2sqrt{3}-8i}{95}$. Are these correct points, and should I continue to solve for the circle?
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– A Piercing Arrow
Jan 2 at 19:19
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@APiercingArrow: Those three points are correct and your first two inversions are correct, but the third inversion is wrong. But ignore all but the first point and its inversion--I edited my answer to show a way using only two points to get your answer. One of those two points is your first point, and I state the second point. Continue from there and ask if you still need help.
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– Rory Daulton
Jan 2 at 23:16
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@APiercingArrow: Those three points are correct and your first two inversions are correct, but the third inversion is wrong. But ignore all but the first point and its inversion--I edited my answer to show a way using only two points to get your answer. One of those two points is your first point, and I state the second point. Continue from there and ask if you still need help.
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– Rory Daulton
Jan 2 at 23:16
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Thank you so much! I got the equation of the circle to be $(x -sqrt{3}/15$ )^2 + (y+i/15)^2 = 13/(225^2)$. With this equation, $a = 5$, $b=3$, $c=15$, and $d$ is not an integer. Where did I go wrong?
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– A Piercing Arrow
Jan 3 at 22:32
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Thank you so much! I got the equation of the circle to be $(x -sqrt{3}/15$ )^2 + (y+i/15)^2 = 13/(225^2)$. With this equation, $a = 5$, $b=3$, $c=15$, and $d$ is not an integer. Where did I go wrong?
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– A Piercing Arrow
Jan 3 at 22:32
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