Let $ \mathbb{R} $ denote the set of all real numbers. Let $ z_1 = 1 + 2i $ and $ z_2 = 3i $ be two complex numbers, where $ i = \sqrt{-1} $. Let
$$
S = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x + iy - z_1| = 2|x + iy - z_2|\}.
$$
Then which of the following statements is (are) TRUE?
Show Hint
- \( S \) is a circle with centre \( \left(-\dfrac{1}{3}, \dfrac{10}{3}\right) \)
- \( S \) is a circle with centre \( \left(\dfrac{1}{3}, \dfrac{8}{3}\right) \)
- \( S \) is a circle with radius \( \dfrac{\sqrt{2}}{3} \)
\( S \) is a circle with radius \( \dfrac{2\sqrt{2}}{3} \)
The Correct Option is A, D
Solution and Explanation
Step 1: Define the given complex numbers and set \( S \)
- We’re given two complex numbers:
- \( z_1 = 1 + 2i \)
- \( z_2 = 3i \), where \( i = \sqrt{-1} \).
- The set \( S \subset \mathbb{R} \times \mathbb{R} \) (the real plane, which we can think of as the complex plane where points \((x, y)\) correspond to complex numbers \( x + iy \)) is defined by: \[ S = \{(x, y) \in \mathbb{R} \times \mathbb{R} \mid |x + iy - z_1| = 2|x + iy - z_2|\}. \]
- Substitute the values of \( z_1 \) and \( z_2 \):
- \( z_1 = 1 + 2i \), so \( x + iy - z_1 = (x + iy) - (1 + 2i) = (x - 1) + i(y - 2) \).
- \( z_2 = 3i \), so \( x + iy - z_2 = (x + iy) - 3i = x + i(y - 3) \).
- The defining equation of \( S \) becomes: \[ |x + iy - z_1| = 2|x + iy - z_2| \] \[ |(x - 1) + i(y - 2)| = 2|x + i(y - 3)|. \]
Step 2: Interpret the equation geometrically
- The modulus \( |x + iy - z_1| \) represents the distance from the point \( (x, y) \) to the point corresponding to \( z_1 \). Since \( z_1 = 1 + 2i \), this is the point \( (1, 2) \) in the plane.
- Similarly, \( |x + iy - z_2| \) is the distance from \( (x, y) \) to the point corresponding to \( z_2 = 3i \), which is \( (0, 3) \).
- The equation \( |x + iy - z_1| = 2|x + iy - z_2| \) means that the distance from \( (x, y) \) to \( (1, 2) \) is twice the distance from \( (x, y) \) to \( (0, 3) \).
Geometrically, the set of points where the distance to one point is a constant multiple (here, 2) of the distance to another point is related to a circle. This is a classic locus problem: the set of points \( P(x, y) \) such that the distance from \( P \) to point \( A \) is \( k \) times the distance from \( P \) to point \( B \) (with \( k \neq 1 \)) forms a circle known as an Apollonius circle.
Step 3: Convert the equation into a workable form
Let’s compute the moduli:
- \( |x + iy - z_1| = |(x - 1) + i(y - 2)| = \sqrt{(x - 1)^2 + (y - 2)^2} \).
- \( |x + iy - z_2| = |x + i(y - 3)| = \sqrt{x^2 + (y - 3)^2} \).
The equation becomes: \[ \sqrt{(x - 1)^2 + (y - 2)^2} = 2 \sqrt{x^2 + (y - 3)^2}. \]
Square both sides to eliminate the square roots: \[ (x - 1)^2 + (y - 2)^2 = 4 \left( x^2 + (y - 3)^2 \right). \]
Expand both sides:
- Left: \( (x - 1)^2 + (y - 2)^2 = (x^2 - 2x + 1) + (y^2 - 4y + 4) = x^2 + y^2 - 2x - 4y + 5 \).
- Right: \( 4 \left( x^2 + (y - 3)^2 \right) = 4 \left( x^2 + (y^2 - 6y + 9) \right) = 4x^2 + 4y^2 - 24y + 36 \).
So: \[ x^2 + y^2 - 2x - 4y + 5 = 4x^2 + 4y^2 - 24y + 36. \]
Move all terms to one side: \[ x^2 + y^2 - 2x - 4y + 5 - 4x^2 - 4y^2 + 24y - 36 = 0. \]
Combine like terms:
- \( x^2 - 4x^2 = -3x^2 \),
- \( y^2 - 4y^2 = -3y^2 \),
- \( -2x \),
- \( -4y + 24y = 20y \),
- \( 5 - 36 = -31 \),
\[ -3x^2 - 3y^2 - 2x + 20y - 31 = 0. \]
Multiply through by \(-1\) for simplicity: \[ 3x^2 + 3y^2 + 2x - 20y + 31 = 0. \]
Divide by 3: \[ x^2 + y^2 + \frac{2}{3}x - \frac{20}{3}y + \frac{31}{3} = 0. \]
Step 4: Complete the square to find the equation of the circle
Rewrite the equation: \[ x^2 + \frac{2}{3}x + y^2 - \frac{20}{3}y = -\frac{31}{3}. \]
- For \( x \)-terms: \( x^2 + \frac{2}{3}x \). The coefficient of \( x \) is \( \frac{2}{3} \), so half of it is \( \frac{1}{3} \), and \( \left(\frac{1}{3}\right)^2 = \frac{1}{9} \). \[ x^2 + \frac{2}{3}x = \left(x + \frac{1}{3}\right)^2 - \frac{1}{9}. \]
- For \( y \)-terms: \( y^2 - \frac{20}{3}y \). The coefficient of \( y \) is \( -\frac{20}{3} \), so half is \( -\frac{10}{3} \), and \( \left(-\frac{10}{3}\right)^2 = \frac{100}{9} \). \[ y^2 - \frac{20}{3}y = \left(y - \frac{10}{3}\right)^2 - \frac{100}{9}. \]
Substitute back: \[ \left(x + \frac{1}{3}\right)^2 - \frac{1}{9} + \left(y - \frac{10}{3}\right)^2 - \frac{100}{9} = -\frac{31}{3}. \]
Combine the constants on the right: \[ -\frac{1}{9} - \frac{100}{9} + \frac{31}{3} = -\frac{1}{9} - \frac{100}{9} + \frac{93}{9} = \frac{-1 - 100 + 93}{9} = \frac{-8}{9}. \]
So: \[ \left(x + \frac{1}{3}\right)^2 + \left(y - \frac{10}{3}\right)^2 = \frac{8}{9}. \]
This is the equation of a circle with:
- Center: \( \left(-\frac{1}{3}, \frac{10}{3}\right) \).
- Radius: \( \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3} \).
Step 5: Evaluate the options
- (A) \( S \) is a circle with center \( \left(-\frac{1}{3}, \frac{10}{3}\right) \): This matches our computed center, so this is true.
- (B) \( S \) is a circle with center \( \left(\frac{1}{3}, \frac{8}{3}\right) \): This does not match the center \( \left(-\frac{1}{3}, \frac{10}{3}\right) \), so this is false.
- (C) \( S \) is a circle with radius \( \frac{\sqrt{2}}{3} \): The radius is \( \frac{2\sqrt{2}}{3} \), not \( \frac{\sqrt{2}}{3} \), so this is false.
- (D) \( S \) is a circle with radius \( \frac{2\sqrt{2}}{3} \): This matches our computed radius, so this is true.
Final Answer: The true statements are:
- (A) \( S \) is a circle with center \( \left(-\frac{1}{3}, \frac{10}{3}\right) \).
- (D) \( S \) is a circle with radius \( \frac{2\sqrt{2}}{3} \).
Thus, the correct options are (A) and (D). % Correct Answer Correct Answer: (A), (D)
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