Question

Convert r = 8cos θ to rectangular form.

Answer 1

Answer:

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

This is a circle with radius $\sqrt{16}=4$ and center $(4,0)$.

Step-by-step explanation:

Recall the following:

$x=r\cos(\theta)$

$y=r\sin(\theta)$

$\frac{y}{x}=\tan(\theta)$

$x^2+y^2=r^2$

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

Multiply both sides by $r$:

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

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 $8x$ 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 $x$ part.

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

$(x+\frac{-8}{2})^2+y^2=0+(-4)^2$

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

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

So our equation is that of a circle with radius $\sqrt{16}=4$ and center $(4,0)$.