How do you convert (x-3)squared +(y+4)squared =25 to a polar equation?

##r=2(3 cos theta – 4 sin theta )##. Graph is inserted.

The conversion formula is

##(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 theta – 4 sin theta )##

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

when ##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]}

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How do you convert (x-3)squared +(y+4)squared =25 to a polar equation?

##r=2(3 cos theta – 4 sin theta )##. Graph is inserted.

The conversion formula is

##(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 theta – 4 sin theta )##

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

when ##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]}

Leave a Comment

Your email address will not be published. Required fields are marked *

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