The given polar equation is: r=6cosθ r = 6 \cos \theta r=6cosθ
Using the relationships x=rcosθ x = r \cos \theta x=rcosθ and r=x2+y2 r = \sqrt{x^2 + y^2} r=x2+y2, we can convert this to rectangular form.
First, multiply both sides by r r r: r2=6rcosθ r^2 = 6r \cos \theta r2=6rcosθ
Since rcosθ=x r \cos \theta = x rcosθ=x: r2=6x r^2 = 6x r2=6x
And since r2=x2+y2 r^2 = x^2 + y^2 r2=x2+y2: x2+y2=6x x^2 + y^2 = 6x x2+y2=6x
Rearrange to get the standard form of a circle: x2+y2−6x=0 x^2 + y^2 - 6x = 0 x2+y2−6x=0
Complete the square for the x x x terms: (x2−6x+9)+y2=9 (x^2 - 6x + 9) + y^2 = 9 (x2−6x+9)+y2=9 (x−3)2+y2=32 (x - 3)^2 + y^2 = 3^2 (x−3)2+y2=32
This is the equation of a circle with center at (3,0) (3, 0) (3,0) and radius 3.
The rectangular form of the given polar equation is: (x−3)2+y2=9 (x - 3)^2 + y^2 = 9 (x−3)2+y2=9
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