How do you find the intersection of $r = 1 - cos(theta)$ , $r^2 = 4cos(theta)$ ?

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Answer 1 · 2016-11-23T11:57:11

$( 2 (sqrt2 - 1), cos^(-1) ( 3 - sqrt 2 ) ) = ( 0.8284, 80.12^o )$

r > 0.

Eliminate r.

$r^2=(1-cos theta )^2=4 cos theta$

Solving,

$cos theta=3-2sqrt2 to theta =arc cos (3-2sqrt2) to r =2((sqrt2-1)$ .

So, the common point is

$( 2 (sqrt2 - 1), cos^(-1) ( 3 - sqrt 2 ) ) = ( 0.8284, 80.12^o )$

How do you find the intersection of r = 1 - cos(theta)r = 1 - cos(theta), r^2 = 4cos(theta)r^2 = 4cos(theta)?