How do you convert #r=12 cos [theta]# to rectangular form?
2 Answers
The equation becomes a circle with radius 6 and centered at
Explanation:
Multiply both sides by r:
Substitute
Subtract 12x from both sides:
Add h^2 to both sides:
Complete the square for #(x - h)^2 = x^2 - 2hx + h^2:
-2hx = -12x
Insert the square term with h = 6 on the left and substitute 6 for h on the right:
Write the y term is standard form:
Explanation:
Multiply the given expression on both sides by r, it becomes
Substitute
The equation now becomes,