Respuesta :
Answer:
[tex](x-4)^2+y^2=16[/tex]
This is a circle with radius [tex]\sqrt{16}=4[/tex] and center [tex](4,0)[/tex].
Step-by-step explanation:
Recall the following:
[tex]x=r\cos(\theta)[/tex]
[tex]y=r\sin(\theta)[/tex]
[tex]\frac{y}{x}=\tan(\theta)[/tex]
[tex]x^2+y^2=r^2[/tex]
We are given [tex]r=8\cos(theta)[/tex].
Multiply both sides by [tex]r[/tex]:
[tex]r(r)=8\cos(\theta)r[/tex]
Reorder on right and simplify on left:
[tex]r^2=8r\cos(theta)[/tex]
We can now make some replacements from our "Recall" section:
[tex]x^2+y^2=8x[/tex]
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 [tex](x-h)^2+(y-k)^2=r^2[/tex].
Subtract [tex]8x[/tex] on both sides:
[tex]x^2-8x+y^2=0[/tex]
[tex]x^2-8x+\text{(blank)}+y^2=0+\text{(blank)}[/tex]
We are going to fill those "blanks" in so that we can complete the square for the [tex]x[/tex] part.
[tex]x^2-8x+(\frac{-8}{2})^2+y^2=0+(\frac{-8}{2})^2[/tex]
[tex](x+\frac{-8}{2})^2+y^2=0+(-4)^2[/tex]
[tex](x-4)^2+y^2=0+16[/tex]
[tex](x-4)^2+y^2=16[/tex]
So our equation is that of a circle with radius [tex]\sqrt{16}=4[/tex] and center [tex](4,0)[/tex].
