SOLUTION: What is the rectangular form of the equation r= -16cos(theta)? The answer choices are: A.(x+8)^2+y^2=64 B. x^2+(y-8)^2=64 and C. (x-8)^2+y^2=64

Algebra ->  Trigonometry-basics -> SOLUTION: What is the rectangular form of the equation r= -16cos(theta)? The answer choices are: A.(x+8)^2+y^2=64 B. x^2+(y-8)^2=64 and C. (x-8)^2+y^2=64       Log On


   

Question 980071: What is the rectangular form of the equation r= -16cos(theta)? The answer choices are: A.(x+8)^2+y^2=64 B. x^2+(y-8)^2=64 and C. (x-8)^2+y^2=64
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Identities Used


r^2 = x^2 + y^2
x = r*cos(theta)


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r= -16cos(theta)


r*r= r*(-16cos(theta)) ... multiply both sides by r


r^2 = -16(rcos(theta))


r^2 = -16x ... replace "rcos(theta)" with x (see above)


x^2 + y^2 = -16x ... replace r^2 with x^2+y^2 (see above)


x^2+16x + y^2 = 0


x^2+16x +64 + y^2 = 0+64 ... see note below


(x^2+16x +64) + y^2 = 64


(x+8)^2 + y^2 = 64


This is a circle with center (-8,0) and radius 8


Note: I took half of the x coefficient (16) to get 8. Then I squared 8 to get 64. I added 64 to both sides. This is all done to complete the square.