How do you convert $1=(x+4)^2+(7y-2)^2$ into polar form?

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Answer 1 · 2016-10-04T09:24:18

$r^2(1+48sin^2theta)+r(8costheta-28sintheta)+19=0$

The relation between rectangular Cartesian coordinates $(x,y)$ and polar coordinates $(r,theta)$ is given by

$x=rcostheta$ , $y=rsintheta$ and $x^2+y^2=r^2$

Hence $y^2+(x-3)^2=9$ can be rewritten as

$(x+4)^2+(7y-2)^2=1$

or $(rcostheta+4)^2+(7rsintheta-2)^2=1$

or $(r^2cos^2theta+8rcostheta+16)+(49r^2sin^2theta-28rsintheta+4)=1$

or $r^2cos^2theta+8rcostheta+49r^2sin^2theta-28rsintheta+20-1=0$

or $r^2(1+48sin^2theta)+r(8costheta-28sintheta)+19=0$

How do you convert 1=(x+4)^2+(7y-2)^21=(x+4)^2+(7y-2)^2 into polar form?