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

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

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Answer 1 · 2016-10-28T17:08:13

$r^{2} - 2 r \sqrt{h^{2} + k^{2}} sin (θ + ϕ_{0}) + h^{2} + k^{2} = a^{2}$ $r^2-2rsqrt(h^2+k^2)sin(theta+phi_0)+h^2+k^2=a^2$
with
$ϕ_{0} = arctan (\frac{h}{k})$ $phi_0=arctan(h/k)$
or also

$r = \sqrt{h^{2} + k^{2}} sin (θ + ϕ_{0}) \pm \sqrt{a^{2} - h^{2} - k^{2} + (h^{2} + k^{2}) {sin (θ + ϕ_{0})}^{2}}$ $r = sqrt[h^2 + k^2] Sin( theta+phi_0) pm sqrt[ a^2 - h^2 - k^2 + (h^2 + k^2) sin(theta+phi_0)^2]$

Explanation:

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

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

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

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Explanation:

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How do you convert (x-h)^2+(y-k)^2=a^2(x−h)2+(y−k)2=a2 (x-h)^2+(y-k)^2=a^2 to polar form?

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