JEE Mains · Maths · STD 11 - 4.1 complex nubers
If the center and radius of the circle \(\left|\frac{z-2}{z-3}\right|=2\) are respectively \((\alpha, \beta)\) and \(\gamma\), then \(3(\alpha+\beta+\gamma)\) is equal to
- A \(11\)
- B \(9\)
- C \(10\)
- D \(12\)
Answer & Solution
Correct Answer
(D) \(12\)
Step-by-step Solution
Detailed explanation
\(\sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2}\) \(=x^2+y^2-4 x+4=4 x^2+4 y^2-24 x+36\) \(=3 x^2+3 y^2-20 x+32=0\) \(=x^2+y^2-\frac{20}{3} x +\frac{32}{3}=0\) \(=(\alpha, \beta)=\left(\frac{10}{3}, 0\right)\) \(\gamma=\sqrt{\frac{100}{9}-\frac{32}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3}\)…
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