Exams Joint Entrance Examination Mathematics Complex Numbers Question 439 Question 439 of 455 Medium MCQ
If the center and radius of the circle \[\left| {{{z - 2} \over {z - 3}}} \right| = 2\] are respectively \[(\alpha,\beta)\] and \[\gamma\], then \[3(\alpha+\beta+\gamma)\] is equal to :
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\[\left| {{{z - 2} \over {z - 3}}} \right| = 2\]
\[
( x − 2 ) 2 + y 2 = 2 ( x − 3 ) 2 + y 2 ⇒ x 2 + y 2 − 4 x + 4 = 4 x 2 + 4 y 2 − 24 x + 36 ⇒ 3 x 2 + 3 y 2 − 20 x + 32 = 0 ⇒ x 2 + y 2 − 20 3 x + 32 3 = 0 ⇒ ( α , β ) = ( 10 3 , 0 ) γ = 100 9 − 32 3 = 4 9 = 2 3 3 ( α + β + γ ) = 3 ( 10 3 + 2 3 ) = 12 \begin{aligned}
& \sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2} \\\\
& \Rightarrow x^2+y^2-4 x+4=4 x^2+4 y^2-24 x+36 \\\\
& \Rightarrow 3 x^2+3 y^2-20 x+32=0 \\\\
& \Rightarrow x^2+y^2-\frac{20}{3} \mathrm{x}+\frac{32}{3}=0 \\\\
& \Rightarrow (\alpha, \beta)=\left(\frac{10}{3}, 0\right) \\\\
& \gamma=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3} \\\\
& 3(\alpha + \beta + \gamma)=3\left(\frac{10}{3}+\frac{2}{3}\right) \\\\
& =12
\end{aligned} ( x − 2 ) 2 + y 2 = 2 ( x − 3 ) 2 + y 2 ⇒ x 2 + y 2 − 4 x + 4 = 4 x 2 + 4 y 2 − 24 x + 36 ⇒ 3 x 2 + 3 y 2 − 20 x + 32 = 0 ⇒ x 2 + y 2 − 3 20 x + 3 32 = 0 ⇒ ( α , β ) = ( 3 10 , 0 ) γ = 9 100 − 3 32 = 9 4 = 3 2 3 ( α + β + γ ) = 3 ( 3 10 + 3 2 ) = 12
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