Question

convert to polar form: (x-3)^2+(y+4)^2=9

Answer 1

Given equation is (x-3)^2+(y+4)^2 = 9

(x-3)^2+(y-(-4))^2 = 3^2

This is a ciircle equation as (x-h)^2+(y-k)^2 = r^2

Center is (h,k) = (3,-4) and radius = 3.

We know that polar coordiinates (r, θ) ,cartisian coordinates (x,y).

x^2+y^2 = r^2

y = r Sinθ

x = r Cosθ

(r Cosθ-3)^2+(r Sinθ+4)^2 = 9

r^2Cos^2θ+9-6rCosθ+r^2Sin^2θ+16+8rSinθ = 9

r^2(Cos^2θ+Sin^2θ)+8rsinθ-6rCosθ+25 = 9

Subtract 9 from each side.

r^2(Cos^2θ+Sin^2θ)+8rsinθ-6rCosθ+25-9 = 9-9

We know that Cos^2θ+Sin^2θ = 1.

r^2(1) +8rsinθ-6rCosθ+16 = 0

Polar form is r^2+8rsinθ-6rCosθ+16 = 0.