Convert (x-1)^2 +y^2=1 to a polar equation
To convert the given equation, (x-1)^2 + y^2 = 1, to a polar equation, we need to express x and y in terms of r and θ.
First, let's rewrite the equation in terms of x and y:
(x-1)^2 + y^2 = 1
Expanding the equation, we get:
x^2 - 2x + 1 + y^2 = 1
Now, we can substitute x and y with their respective polar representations:
x = rcos(θ)
y = rsin(θ)
Replacing x and y in the equation, we get:
(r*cos(θ))^2 - 2(r*cos(θ)) + 1 + (r*sin(θ))^2 = 1
Simplifying, we have:
r^2*cos^2(θ) - 2r*cos(θ) + 1 + r^2*sin^2(θ) = 1
Since we know that cos^2(θ) + sin^2(θ) = 1, we can substitute this expression into the equation:
r^2*cos^2(θ) + r^2*sin^2(θ) = 2r*cos(θ)
Simplifying further:
r^2 * (cos^2(θ) + sin^2(θ)) = 2r*cos(θ)
Using the identity cos^2(θ) + sin^2(θ) = 1, we get:
r^2 = 2r*cos(θ)
Finally, we have obtained the polar equation:
r^2 - 2r*cos(θ) = 0
just crank it out.
(rcosθ - 1)^2 + r^2 sin^2θ = 1
r^2 cos^2θ - 2rcosθ + r^2 sin^2θ = 1
r^2 - 2rcosθ - 1 = 0
see the graph, which is, as expected, a circle with center at (1,0)
https://www.wolframalpha.com/input/?i=r%5E2+-+2rcos%CE%B8+-+1+%3D+0