# graph in the xy plane is

• Apr 14th 2010, 01:26 PM
bigwave
graph in the xy plane is
The graph in the xy-plane represented by
$\displaystyle x=3+2\sin{\theta}$
$\displaystyle y=2\cos{\theta}-1$
for $\displaystyle -\pi\le\theta\le\pi$

the answer is a circle with the equation
$\displaystyle (x-3)^2+(y+1)^2=4$

but got stuck on how this was derived especially from the trig functions(Hi)
• Apr 14th 2010, 02:10 PM
skeeter
Quote:

Originally Posted by bigwave
The graph in the xy-plane represented by
$\displaystyle x=3+2\sin{\theta}$
$\displaystyle y=2\cos{\theta}-1$
for $\displaystyle -\pi\le\theta\le\pi$

the answer is a circle with the equation
$\displaystyle (x-3)^2+(y+1)^2=4$

but got stuck on how this was derived especially from the trig functions(Hi)

$\displaystyle x^2 = 9 + 12\sin{t} + 4\sin^2{t}$

$\displaystyle y^2 = 4\cos^2{t} - 4\cos{t} + 1$

-----------------------------------

$\displaystyle x^2 + y^2 = 9 + 4 + 1 + 12\sin{t} - 4\cos{t}$

$\displaystyle x^2 + y^2 = 14 + 12\sin{t} - 4\cos{t}$

$\displaystyle x^2 + y^2 = 18 + 12\sin{t} - 4 - 4\cos{t}$

$\displaystyle x^2 + y^2 = 6(3 + 2\sin{t}) + 2 - 4\cos{t} - 6$

$\displaystyle x^2 + y^2 = 6(3 + 2\sin{t}) - 2(2\cos{t} - 1) - 6$

$\displaystyle x^2 + y^2 = 6x - 2y - 6$

$\displaystyle x^2 - 6x + y^2 + 2y = -6$

$\displaystyle x^2 - 6x + 9 + y^2 + 2y + 1 = -6 + 9 + 1$

$\displaystyle (x - 3)^2 + (y + 1)^2 = 4$