Given:
∣z−3z−2∣=2
Let z=x+iy, then we have
∣x−3+iyx−2+iy∣=2
⇒(x−2)2+y2=2(x−3)2+y2
⇒x2+y2−4x+4=4x2+4y2−24x+36
⇒3x2+3y2−20x+32=0
⇒x2+y2−320x+332=0
So, (α,β)≡(310,0)
And,
γ=9100−332=94=32
Therefore,
3(α+β+γ)=3(310+32)=12
JEE Main 2023 — Mathematics Algebra
If the center and radius of the circle ∣z−3z−2∣=2 are respectively (α,β) and γ, then 3(α+β+γ) is equal to
Held on 1 Feb 2023 · Verified 6 Jul 2026.
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