Convert $(x-3)^2 + y^2 = 49$ to polar form.
Convert $(x-3)^2 + y^2 = 49$ to polar form.
Applying $x=rcos(theta)$ and $y=rsin(theta)$, I get
$x^2 - 6x + 9 + y^2 = 49$
$r^2-6x = 40$
$r^2-6rcos(theta) = 40$
$r(r-6cos(theta)) = 40$
This definitely looks wrong. What am I doing wrong?
algebra-precalculus polar-coordinates
asked Nov 28 at 22:35
Axion004
217212
add a comment |
Convert $(x-3)^2 + y^2 = 49$ to polar form.
Applying $x=rcos(theta)$ and $y=rsin(theta)$, I get
$x^2 - 6x + 9 + y^2 = 49$
$r^2-6x = 40$
$r^2-6rcos(theta) = 40$
$r(r-6cos(theta)) = 40$
This definitely looks wrong. What am I doing wrong?
algebra-precalculus polar-coordinates
asked Nov 28 at 22:35
Axion004
217212
2
Try $x = 7costheta+3, ; y=7sintheta$.
– MisterRiemann
Nov 28 at 22:45
How do you know that $x=7cos(theta)+3$? How did you compute that?
– Axion004
Nov 28 at 22:48
The equation describes the circle centered at $(3,0)$ (which is why $3$ appears in $x$) and with radius $7$ (which is why $7$ appears in front of both $cos$ and $sin$). Judging by the newly posted answer though, I may have misunderstood the exercise.
– MisterRiemann
Nov 28 at 22:49
I see, I got it. Thank you.
– Axion004
Nov 28 at 22:51
add a comment |
Convert $(x-3)^2 + y^2 = 49$ to polar form.
Applying $x=rcos(theta)$ and $y=rsin(theta)$, I get
$x^2 - 6x + 9 + y^2 = 49$
$r^2-6x = 40$
$r^2-6rcos(theta) = 40$
$r(r-6cos(theta)) = 40$
This definitely looks wrong. What am I doing wrong?
algebra-precalculus polar-coordinates
asked Nov 28 at 22:35
Axion004
217212
Convert $(x-3)^2 + y^2 = 49$ to polar form.
Applying $x=rcos(theta)$ and $y=rsin(theta)$, I get
$x^2 - 6x + 9 + y^2 = 49$
$r^2-6x = 40$
$r^2-6rcos(theta) = 40$
$r(r-6cos(theta)) = 40$
This definitely looks wrong. What am I doing wrong?
algebra-precalculus polar-coordinates
algebra-precalculus polar-coordinates
asked Nov 28 at 22:35
Axion004
217212
asked Nov 28 at 22:35
Axion004
217212
asked Nov 28 at 22:35
Axion004
217212
asked Nov 28 at 22:35
Axion004
217212
asked Nov 28 at 22:35
Axion004
217212
217212
2
Try $x = 7costheta+3, ; y=7sintheta$.
– MisterRiemann
Nov 28 at 22:45
How do you know that $x=7cos(theta)+3$? How did you compute that?
– Axion004
Nov 28 at 22:48
The equation describes the circle centered at $(3,0)$ (which is why $3$ appears in $x$) and with radius $7$ (which is why $7$ appears in front of both $cos$ and $sin$). Judging by the newly posted answer though, I may have misunderstood the exercise.
– MisterRiemann
Nov 28 at 22:49
I see, I got it. Thank you.
– Axion004
Nov 28 at 22:51
add a comment |
2
Try $x = 7costheta+3, ; y=7sintheta$.
– MisterRiemann
Nov 28 at 22:45
How do you know that $x=7cos(theta)+3$? How did you compute that?
– Axion004
Nov 28 at 22:48
The equation describes the circle centered at $(3,0)$ (which is why $3$ appears in $x$) and with radius $7$ (which is why $7$ appears in front of both $cos$ and $sin$). Judging by the newly posted answer though, I may have misunderstood the exercise.
– MisterRiemann
Nov 28 at 22:49
I see, I got it. Thank you.
– Axion004
Nov 28 at 22:51
2
2
Try $x = 7costheta+3, ; y=7sintheta$.
– MisterRiemann
Nov 28 at 22:45
Try $x = 7costheta+3, ; y=7sintheta$.
– MisterRiemann
Nov 28 at 22:45
How do you know that $x=7cos(theta)+3$? How did you compute that?
– Axion004
Nov 28 at 22:48
How do you know that $x=7cos(theta)+3$? How did you compute that?
– Axion004
Nov 28 at 22:48
The equation describes the circle centered at $(3,0)$ (which is why $3$ appears in $x$) and with radius $7$ (which is why $7$ appears in front of both $cos$ and $sin$). Judging by the newly posted answer though, I may have misunderstood the exercise.
– MisterRiemann
Nov 28 at 22:49
The equation describes the circle centered at $(3,0)$ (which is why $3$ appears in $x$) and with radius $7$ (which is why $7$ appears in front of both $cos$ and $sin$). Judging by the newly posted answer though, I may have misunderstood the exercise.
– MisterRiemann
Nov 28 at 22:49
I see, I got it. Thank you.
– Axion004
Nov 28 at 22:51
I see, I got it. Thank you.
– Axion004
Nov 28 at 22:51
add a comment |
2 Answers
2
active
oldest
votes
That's correct indeed we have
$$x^2 - 6x + 9 + y^2 = 49 implies r^2-6rcos theta=40$$
that is the polar form of a circle centered at $(3,0)$ and radius $7$.
If we refer the origin at $(3,0)$ we obtain indeed $r=7$.
answered Nov 28 at 22:48
gimusi
1
add a comment |
To convert to polar form, set x=rcos($theta$) and $y=rsin(theta)=$ So we get $(rcos(theta)-3)^2$+$r^2sin^2(theta)=49$, expanding gives you $r^2cos(theta)^2-6rcos(theta)+9+r^2sin^2(theta)=49$ then using the trig identity $sin^2(theta)+cos^2(theta)=1$ yields $r^2-6rcos(theta)=40$
answered Nov 28 at 23:00
maths researcher
458
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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oldest
votes
That's correct indeed we have
$$x^2 - 6x + 9 + y^2 = 49 implies r^2-6rcos theta=40$$
that is the polar form of a circle centered at $(3,0)$ and radius $7$.
If we refer the origin at $(3,0)$ we obtain indeed $r=7$.
answered Nov 28 at 22:48
gimusi
1
add a comment |
That's correct indeed we have
$$x^2 - 6x + 9 + y^2 = 49 implies r^2-6rcos theta=40$$
that is the polar form of a circle centered at $(3,0)$ and radius $7$.
If we refer the origin at $(3,0)$ we obtain indeed $r=7$.
answered Nov 28 at 22:48
gimusi
1
add a comment |
That's correct indeed we have
$$x^2 - 6x + 9 + y^2 = 49 implies r^2-6rcos theta=40$$
that is the polar form of a circle centered at $(3,0)$ and radius $7$.
If we refer the origin at $(3,0)$ we obtain indeed $r=7$.
answered Nov 28 at 22:48
gimusi
1
That's correct indeed we have
$$x^2 - 6x + 9 + y^2 = 49 implies r^2-6rcos theta=40$$
that is the polar form of a circle centered at $(3,0)$ and radius $7$.
If we refer the origin at $(3,0)$ we obtain indeed $r=7$.
answered Nov 28 at 22:48
gimusi
1
answered Nov 28 at 22:48
gimusi
1
answered Nov 28 at 22:48
gimusi
1
answered Nov 28 at 22:48
gimusi
1
1
add a comment |
add a comment |
To convert to polar form, set x=rcos($theta$) and $y=rsin(theta)=$ So we get $(rcos(theta)-3)^2$+$r^2sin^2(theta)=49$, expanding gives you $r^2cos(theta)^2-6rcos(theta)+9+r^2sin^2(theta)=49$ then using the trig identity $sin^2(theta)+cos^2(theta)=1$ yields $r^2-6rcos(theta)=40$
answered Nov 28 at 23:00
maths researcher
458
add a comment |
To convert to polar form, set x=rcos($theta$) and $y=rsin(theta)=$ So we get $(rcos(theta)-3)^2$+$r^2sin^2(theta)=49$, expanding gives you $r^2cos(theta)^2-6rcos(theta)+9+r^2sin^2(theta)=49$ then using the trig identity $sin^2(theta)+cos^2(theta)=1$ yields $r^2-6rcos(theta)=40$
answered Nov 28 at 23:00
maths researcher
458
add a comment |
To convert to polar form, set x=rcos($theta$) and $y=rsin(theta)=$ So we get $(rcos(theta)-3)^2$+$r^2sin^2(theta)=49$, expanding gives you $r^2cos(theta)^2-6rcos(theta)+9+r^2sin^2(theta)=49$ then using the trig identity $sin^2(theta)+cos^2(theta)=1$ yields $r^2-6rcos(theta)=40$
answered Nov 28 at 23:00
maths researcher
458
To convert to polar form, set x=rcos($theta$) and $y=rsin(theta)=$ So we get $(rcos(theta)-3)^2$+$r^2sin^2(theta)=49$, expanding gives you $r^2cos(theta)^2-6rcos(theta)+9+r^2sin^2(theta)=49$ then using the trig identity $sin^2(theta)+cos^2(theta)=1$ yields $r^2-6rcos(theta)=40$
answered Nov 28 at 23:00
maths researcher
458
answered Nov 28 at 23:00
maths researcher
458
answered Nov 28 at 23:00
maths researcher
458
answered Nov 28 at 23:00
maths researcher
458
458
add a comment |
add a comment |
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2
Try $x = 7costheta+3, ; y=7sintheta$.
– MisterRiemann
Nov 28 at 22:45
How do you know that $x=7cos(theta)+3$? How did you compute that?
– Axion004
Nov 28 at 22:48
The equation describes the circle centered at $(3,0)$ (which is why $3$ appears in $x$) and with radius $7$ (which is why $7$ appears in front of both $cos$ and $sin$). Judging by the newly posted answer though, I may have misunderstood the exercise.
– MisterRiemann
Nov 28 at 22:49
I see, I got it. Thank you.
– Axion004
Nov 28 at 22:51