Prove that `|Z-Z_1|^2+|Z-Z_2|^2=a` will represent a real circle [with center `(|Z_1+Z_2|^//2+)` ] on the Argand plane if `2ageq|Z_1-Z_1|^2`

Options

A. `k lt |z_(1)-z_(2)|^(2)`
B. `k = |z_(1)-z_(2)|^(2)`
C. `k ge 1/2|z_(1)-z_(2)|^(2)`
D. `k lt 1/2|z_(1)-z_(2)|^(2)`

Answer

C

Explanation

We have,
`|z-z_(1)|^(2)+|z-z_(2)|^(2)=k`
`rArr |z|^(2)+|z_(1)|^(2)-2"Re"(zbarz_(1))+|z|^(2)+|z_(2)|^(2)-2"Re"(zbarz_(2))=k`
`rArr |z|^(2)-2"Re"(z(barz_(1)+barz_(2))/2) =1/2{k-|z_(1)|^(2)-|z_(2)|^(2)}`
`rArr |z|^(2)-2"Re"(z(barz_(1)+barz_(2))/2)+|(z_(1)+z_(2))/(2)|^(2)=1/2{k-|z_(1)|^(2)-|z_(2)|^(2)}+|(z_(1)+z_(2))/(2)|^(2)`
`rArr |z-(z_(1)+z_(2))/(2)|^(2)=1/4{2k-2|z_(1)|^(2)-2|z_(2)|^(2)+|z_(1)|^(2)+|z_(2)|^(2)+2"Re"(z_(1)barz_(2))}`
`rArr |z-(z_(1)+z_(2))/(2)|^(2)=1/2sqrt(2k-|z_(1)-z_(2)|^(2))`
Clearly, it represent a circle having center at `(z_(1)+z_(2))/(2)` and radius `1/2sqrt(2k-|z_(1)-z_(2)|^(2))`. For the circle to exist, we must have
`2k-|z_(1)-z_(2)|^(2) ge 0 rArr k ge 1/2|z_(1)-z_(2)|^(2)`