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

Question

1138 views around the world

You can reuse this answer
Creative Commons License

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#