# Replace the Cartesian equation $(x-5)^2 + y^2 = 25$ by an equivalent polar equation.

Replace the Cartesian equation $(x-5)^2 + y^2 = 25$ by an equivalent polar equation.

Let $t= \theta$, $r=5$, $x=r\cos t$, $y=r\sin t$.

I began with $x=5\cos t-5=5(\cos t-1)$ and $y=5\sin t$. Is that the correct procedure?

Then for $y=mx$, $$5\sin t=m(5\cos t-5)$$ which is $$m=(5\sin t)/(5(\cos t-1))$$ so $$m=(\sin t)/(\cos t-1).$$ Does this mean that the polar equation is $r=(\sin t)/(\cos t-1)$? Or is there more to this problem? Is there some other method I should use to obtain the polar equation?

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No, not quite. Though the radius of the circle defined by $$\tag{1}(x-5)^2+y^2=25$$ has value $5$, this is not the "$r$" defining the polar coordinates of a point on the circle (it would be if the circle were centered at the origin).
From the definition of polar coordinates, you simply have to replace in $(1)$ the expression "$x$" with "$r\cos t$" and the expression "$y$" with "$r\sin t$" This would give you $$\tag{2} (r\cos t -5)^2 +(r\sin t)^2=25.$$ Now you should simplify: equation $(2)$ is equivalent to $$r^2\cos^2 t -10r\cos t+25 +r^2\sin^2 t =25,$$ or $$r^2(\sin^2 t+\cos^2 t)=10r\cos t$$ or $$\tag{3} r^2=10r\cos t.$$ One can show that the graph of the polar equation $(3)$ is the same as the graph of the polar equation $(4)$ below (it would be a good exercise to verify this). $$\tag{4} r=10\cos t.$$ And that's all there is to it.
I'm not sure why you are considering "then for $y=mx$"...