JEE Advanced · Mathematics · 3. Complex Numbers
Let \(\mathbb{R}\) denote the set of all real numbers. Let \(z_1=1+2 i\) and \(z_2=3 i\) be two complex numbers, where \(i=\sqrt{-1}\).
Let \(S=\{(x, y) \in \mathbb{R} \times \mathbb{R}:\left|x+i y-z_1\right|=2\)\(\left|x+i y-z_2\right|\}\).
Then which of the following statements is (are) TRUE?
- A \(S\) is a circle with centre \(\left(-\frac{1}{3}, \frac{10}{3}\right)\)
- B \(S\) is a circle with centre \(\left(\frac{1}{3}, \frac{8}{3}\right)\)
- C \(S \text { is a circle with radius } \frac{\sqrt{2}}{3}\)
- D \(S \text { is a circle with radius } \frac{2 \sqrt{2}}{3}\)
Answer & Solution
Correct Answer
(D) \(S \text { is a circle with radius } \frac{2 \sqrt{2}}{3}\)
Step-by-step Solution
Detailed explanation
\(|x+i y-1-2 i|=2|x+i y-3 i| \)
\( \Rightarrow (x-1)^2+(y-2)^2=4\left(x^2+(y-3)^2\right) \)
\( \Rightarrow 3 x^2+3 y^2+2 x-20 y+31=0 \)
\( \Rightarrow x^2+y^2+\frac{2 x}{3}-\frac{20 y}{3}+\frac{31}{3}=0\)
\( \therefore\) S is a circle with centre \(\left(-\frac{1}{3}, \frac{10}{3}\right)\) and radius \(=\sqrt{\frac{1}{9}+\frac{100}{9}-\frac{31}{3}}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}\)
\( \Rightarrow (x-1)^2+(y-2)^2=4\left(x^2+(y-3)^2\right) \)
\( \Rightarrow 3 x^2+3 y^2+2 x-20 y+31=0 \)
\( \Rightarrow x^2+y^2+\frac{2 x}{3}-\frac{20 y}{3}+\frac{31}{3}=0\)
\( \therefore\) S is a circle with centre \(\left(-\frac{1}{3}, \frac{10}{3}\right)\) and radius \(=\sqrt{\frac{1}{9}+\frac{100}{9}-\frac{31}{3}}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}\)
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