How do you convert # (x-h)^2+(y-k)^2=a^2# to polar form?

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Answer 1 · 2016-10-28T17:08:13

#r^2-2rsqrt(h^2+k^2)sin(theta+phi_0)+h^2+k^2=a^2#
with
#phi_0=arctan(h/k)#
or also

#r = sqrt[h^2 + k^2] Sin( theta+phi_0) pm sqrt[ a^2 - h^2 - k^2 + (h^2 + k^2) sin(theta+phi_0)^2]#

The pass equations are

#{(x=rcostheta),(y=rsintheta):}#

then

#(rcostheta-h)^2+(rsintheta-k)^2 =a^2# or

#r^2-2rhcostheta-2rksintheta+h^2+k^2=a^2# or

#r^2-2r(hcostheta+ksintheta)+h^2+k^2=a^2#

Making now

#h/k=tanphi_0# we have

#r^2-2rk(tanphi_0costheta+sintheta)+h^2+k^2=a^2# or

#r^2-2r(k/cosphi_0) (sinphi_0costheta+cosphi_0sintheta)+h^2+k^2= a^2#

but #cosphi_0=k/sqrt(h^2+k^2)# so

#r^2-2rsqrt(h^2+k^2)(sinphi_0costheta+cosphi_0sintheta)+h^2+k^2=a^2#

Finally

#r^2-2rsqrt(h^2+k^2)sin(theta+phi_0)+h^2+k^2=a^2# or also

#r = sqrt[h^2 + k^2] Sin( theta+phi_0) pm sqrt[ a^2 - h^2 - k^2 + (h^2 + k^2) sin(theta+phi_0)^2]#