Calc - Polar Coordinates?

Question

I have a test on Monday on this stuff and these were the questions on last years exam. Can someone help me with these please?! Thanks!

1.) A circle C has center at the origin and radius 7. Another circle K has a diameter with one end at the origin and the other end at the point (0,19). The circles C and K intersect in two points. Let P be the point of intersection of C and K which lies in the first quadrant. Let (r, theta ) be the polar coordinates of P, chosen so that r is positive and 0 is less than or eq to theta whis is less than or eq to 2. Find r and theta.

2.) A curve in polar coordinates is given by: r = 12 + 3 cos theta.
Point P is at theta = (27pi/24)
a.) Find polar coordinate r for P, with r > 0, pi < theta < (3pi/2)
b.) Find cartesian coordinates for point P.
c.) How may times does the curve pass through the origin when 0< theta , 2pi ?

Answer 1

1) Find the intersection of the circles in the first quadrant.

r = 7
r = 19sinθ

19sinθ = 7
sinθ = 7/19
θ = arcsin(7/19)

The intersection of the circles in the first quadrant is:
P(r, θ) = P[7, arcsin(7/19)]
______________

2.) A curve in polar coordinates is given by:
r = 12 + 3cosθ
P(r, θ) = P(r, 27π/24) = P(r, 9π/8)

cos(9π/8) = -cos(π/8) = -(1/2)√(2 + √2)

a)
r = 12 + 3cosθ = 12 + 3cos(9π/8)
r = 12 - (3/2)√(2 + √2)

P(r, θ) = P(r, 27π/24) = P[12 - (3/2)√(2 + √2), 9π/8]
__________

b)
x = rcosθ
y = rsinθ
x² + y² = r²
____________

c) Never
r = 12 + 3cosθ = 0
3cosθ = -12
cosθ = -4

No solution.

Answer 2

x^2+y^2=49 (C)
y/x *(y-19)/x= -1 so x^2+y^2 -19y=0
so subtract
19y=49 and y = 49/19 x^2+(49/19)^2=49

Check numbers
r=12+3 cos t
r= 12+3 cos (9pi/8) so r=9.23 and t=3.534 rad
x= -8.53 and y = -3.53
Check calculations
c)12+3cos t = 0 so cos t = -4 impossible ( never)
delete 1) it has a lot of operational mistakes
x^2+(49/19)^2 = 49 so x= 6.51 and y/x= and t=0.3773 rad
r=7

Answer 3

r = sqrt(x^2 + y^2) = sqrt((-0.5)^2 + (-0.5)^2) = sqrt(0.5), so which you have that incredible. As for theta = arctan(y/x) = arctan(a million), which might many times be pi/4, yet considering the fact that the two x and y are unfavourable all of us be attentive to that theta is interior the third quadrant, and for this reason we end that theta = 5pi/4. So (-0.5, -0.5) in polar form is (sqrt(0.5), 5pi/4).