Convert r = 8cos θ to rectangular form.

Question

asked Mar 21, 2021 56.2k views

1 Answer

Geauser · Answer 1 · 2021-03-26T08:28:17+0000

Answer:

(x-4)^2+y^2=16

This is a circle with radius
√(16)=4 and center
(4,0) .

Explanation:

Recall the following:

$x=r\cos(\theta)$

$y=r\sin(\theta)$

$(y)/(x)=\tan(\theta)$

x^2+y^2=r^2

We are given
$r=8\cos(theta)$ .

Multiply both sides by
:

$r(r)=8\cos(\theta)r$

Reorder on right and simplify on left:

$r^2=8r\cos(theta)$

We can now make some replacements from our "Recall" section:

x^2+y^2=8x

We can probably leave like this, but sometimes we are required to put in a more identifying form.

This is a circle. I'm going to write in the form
(x-h)^2+(y-k)^2=r^2 .

Subtract
on both sides:

x^2-8x+y^2=0

$x^2-8x+\text{(blank)}+y^2=0+\text{(blank)}$

We are going to fill those "blanks" in so that we can complete the square for the
part.

x^2-8x+((-8)/(2))^2+y^2=0+((-8)/(2))^2

(x+(-8)/(2))^2+y^2=0+(-4)^2

(x-4)^2+y^2=0+16

(x-4)^2+y^2=16

So our equation is that of a circle with radius
√(16)=4 and center
(4,0) .