Question

What is the parametric equation for 9 =(x+3)^2+(y-1)^2????

Answer 1

this is circle of radius 3, so lets use these substitutions for x and y:
x = 3cost
y = 3sint
plugging that in we get:
\[9 = (3\cos(t)+3)^{2} +(3\sin(t)-1)^{2}\]
Simplifying gives us:
\[9 = 9\cos^{2}(t)+18\cos(t)+9+9\sin^{2}(t)-6\sin(t)+1\]
\[9 = 19+18\cos(t)-6\sin(t) \Rightarrow -10 = 18\cos(t)-6\sin(t) \Rightarrow -5 = 9\cos(t)-3\sin(t)\]

Answer 2

how did you get a radius of 3 ?

Answer 3

equation of a circle is:
\[(x-h)^{2}+(y-k)^{2} = r^{2}\]
(h,k) is the center of the circle, r is the radius.
Thus:
\[r^{2} = 9 \Rightarrow r = 3\]

Answer 4

If you need anything else explained let me know :)

Answer 5

gotcha......th

Answer 6

gotcha......th

Answer 7

gotcha......th

Answer 8

gotcha......th

Answer 9

your simplifying is a little hard to understand...

Answer 10

sry bout that, one key thing is:
\[9\cos^{2}+9\sin^{2} = 9(\cos^{2}+\sin^{2}) = 9(1) = 9\]
( took out the t cause im lazy >.< imagine its there >.>)