Question

find the locus of z = x + iy if

Answer 1

Dear Student ,

Please find below the solution to the asked query :

Let z=x+iy∣∣z−1∣∣+∣∣z−2∣∣=1∣∣x+iy−1∣∣+∣∣x+iy−2∣∣=1(x−1)2+y2‾‾‾‾‾‾‾‾‾‾‾‾√+(x−2)2+y2‾‾‾‾‾‾‾‾‾‾‾‾√=1(x−1)2+y2‾‾‾‾‾‾‾‾‾‾‾‾√ =1−(x−2)2+y2‾‾‾‾‾‾‾‾‾‾‾‾√Squaring on both sides , we get(x−1)2+y2=1+(x−2)2+y2−2((x−2)2+y2‾‾‾‾‾‾‾‾‾‾‾‾√)x2+1−2x+y2=1+x2+4−4x+y2−2((x−2)2+y2‾‾‾‾‾‾‾‾‾‾‾‾√)−2x=4−4x−2((x−2)2+y2‾‾‾‾‾‾‾‾‾‾‾‾√)2x−4=−2((x−2)2+y2‾‾‾‾‾‾‾‾‾‾‾‾√)Squaring on both sides , we get(2x−4)2=4((x−2)2+y2)4x2+16−16x=4(x2+4−4x+y2)4(x2+4−4x)=4(x2+4−4x+y2)y2=0   is the locus .

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