Convert r = 8cos θ to rectangular form.

This is a circle with radius and center .

Step-by-step explanation:

Recall the following:

We are given .

Multiply both sides by :

Reorder on right and simplify on left:

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

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 .

Subtract on both sides:

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

So our equation is that of a circle with radius and center .

here is a big fat

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)