identify the curve by finding a cartesian equation for the curve. r=2sin(theta) + 2cos(theta)?

Answer 1

r = 2a cosθ + 2b sinθ

is actually a circle passing through the origin... with center at (a,b)

r² = 2a rcosθ + 2b rsinθ

x² + y² = 2ax + 2by
x² - 2ax + a² + y² - 2by + b² = a² + b²
(x - a)² + (y - b)² = a² + b²


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your circle has center at (1,1) in cartesian and has radius √2

Answer 2

r = 2sinθ + 2cosθ
Multiply both sides by r,
r^2 = 2rsinθ + 2rcosθ
x^2+y^2 = 2y+2x, since y = rsinθ, and x = rcosθ
=> x^2-2x+1 + y^2-2y+1 = 2
=>(x-1)^2 + (y-1)^2 = 2
So the curve represents a circle with center at (1, 1) and radius sqrt(2).

Answer 3

just looking at the graph, it's a circle, center at (1,1), radius √2, but I can't prove it yet.

(x-1)² + (y-1)² = 2

ah! since r² = x² + y², and x = rcos Θ and y = rsin Θ, we have

r = 2sin(Θ) + 2cos(Θ)
r² = 2rsin Θ + 2rcos Θ
x² + y² = 2y + 2x
x² - 2x + y² - 2y = 0
x² - 2x + 1 + y² - 2y + 1 = 2
(x-1)² + (y-1)² = 2

should finish my coffee before tackling trig.