If the center and radius of the circle |(z-2/z-3)|=2 are respectively (α, β) and γ, then 3(α+β+γ) is equal to

Question

If the center and radius of the circle |(z-2/z-3)|=2 are respectively (α, β) and &gamma;, then 3(α+β+&gamma;) is equal to (A) 12 (B) 11 (C) 10 (D) 9

Tardigrade Admin · Accepted Answer

√(x-2)2+y2=2 √(x-3)2+y2 =x2+y2-4 x+4=4 x2+4 y2-24 x+36 =3 x2+3 y2-20 x+32=0 =x2+y2-(20/3) x+(32/3)=0 =(α, β)=((10/3), 0) &gamma;=√(100/9)-(32/3)=√(4/9)=(2/3) 3(α, β, &gamma;)=3((10/3)+(2/3)) =12

Q. 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

12

11

10

9

(x−2)2+y2​=2(x−3)2+y2​ =x2+y2−4x+4=4x2+4y2−24x+36 =3x2+3y2−20x+32=0 =x2+y2−320​x+332​=0 =(α,β)=(310​,0) γ=9100​−332​​=94​​=32​ 3(α,β,γ)=3(310​+32​) =12

Q. If the center and radius of the circle $∣ ∣ \frac{z - 2}{z - 3} ∣ ∣ = 2$ are respectively $(α, β)$ and $γ$ , then $3 (α + β + γ)$ is equal to