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#