Question #3583d

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Answer 1 · 2016-11-27T07:29:22

$r = 2 (3 cos θ - 4 sin θ)$ $r=2(3 cos theta - 4 sin theta )$ . Graph is inserted.

The conversion formula is

$(x, y) = r (cos θ, sin θ)$ $(x, y)=r(cos theta, sin theta)$

So, the polar equation is

(r cos theta- 3 )^2+(r sin theta + 4 )^2=25#.

Expanding and simplifying,

$r = 2 (3 cos θ - 4 sin θ)$ $r=2(3 cos theta - 4 sin theta )$

The circle passes through the pole (r = 0 ) ,

when $θ = {tan}^{- 1} (\frac{3}{4}) = {36.87}^{o}$ $theta =tan^(-1)(3/4)=36.87^o$ and also when #theta =

216.87^o#.

This interpretation of reaching the pole is important, in the polar

frame. This discloses the directions of entry into, and exit from, the

pole.

graph{(x-3)^2+(y+4)^2-25=0 [-20, 20, -10, 10]}