How do you convert r=12 cos [theta] to rectangular form?

2 Answers
Nov 25, 2016

The equation becomes a circle with radius 6 and centered at (6, 0): (x - 6)^2 + (y - 0)^2 = 6^2

Explanation:

Multiply both sides by r:

r^2 = 12rcos(theta)

Substitute x^2 + y^2 for r^2 and x for rcos(theta)

x^2 + y^2 = 12x

Subtract 12x from both sides:

x^2 - 12x + y^2 = 0

Add h^2 to both sides:

x^2 - 12x + h^2 + y^2 = h^2

Complete the square for #(x - h)^2 = x^2 - 2hx + h^2:

-2hx = -12x

h = 6

Insert the square term with h = 6 on the left and substitute 6 for h on the right:

(x - 6)^2 + y^2 = 6^2

Write the y term is standard form:

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

Nov 25, 2016

x^2+y^2-12x= 0

Explanation:

Multiply the given expression on both sides by r, it becomes r^2= 12 r cos theta

Substitute rcos theta =x and r^2= x^2+y^2

The equation now becomes, x^2+y^2 = 12x
x^2 +y^2 -12x=0