Write the polar equation in rectangular form. r = 12 sin theta

Write the rectangular equation (x + 7)^2 + y^2 = 49 in polar form

I mixed up x and y in the 1st one!

x^2+y^2 = 12y
x^2 - 12y + y^2 = 0
x^2 + (y-6)^2 = 36

Also, the 2nd one could just as well have been written r = 14cosθ. It would just draw the circle from the left, rather than from the right.

r = 12 sinθ

r^2 = 12r sinθ
x^2+y^2 = 12x
x^2 - 12x + y^2 = 0
(x-6)^2 + y^2 = 36

This one is just the reverse process:

(x+7)^2 + y^2 = 49
x^2 + 14x + y^2 = 0
x^2 + y^2 = -14x
r^2 = -14r cosθ
r = -14 cosθ

thank you!

To convert the polar equation r = 12 sin theta into rectangular form, we can make use of the relationship between polar and rectangular coordinates:

x = r * cos(theta)
y = r * sin(theta)

Substituting r = 12 sin theta into these equations, we get:

x = 12 sin(theta) * cos(theta)
y = 12 sin^2 (theta)

Thus, the rectangular form of the polar equation r = 12 sin theta is:

x = 12 sin(theta) * cos(theta)
y = 12 sin^2 (theta)

To convert the rectangular equation (x + 7)^2 + y^2 = 49 into polar form, we can use the following relationships:

x = r * cos(theta)
y = r * sin(theta)

Substituting these into the equation, we have:

(r * cos(theta) + 7)^2 + (r * sin(theta))^2 = 49

Expanding and simplifying, we get:

r^2 * cos^2(theta) + 14r * cos(theta) + 49 + r^2 * sin^2(theta) = 49

Since cos^2(theta) + sin^2(theta) = 1, we can rewrite the equation as:

r^2 + 14r * cos(theta) = 0

This is the polar form of the given rectangular equation.