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If then locus of is circle then distance from the center of circle to the point (3, 4) is
PS - Question is not original.
The answer is 5.
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1 solution
Let z = a + bi and |z| = sqrt(a^2 + b^2) for the magnitude of this complex number. Then the above absolute value equation becomes:
|2z - 1| = |z - 2|;
or |2*(a+bi) - 1| = |(a+bi) - 2|;
or |(2a-1) + 2bi| = |(a-2) + bi|;
or sqrt[(2a-1)^2 + (2b)^2] = sqrt[(a-2)^2 + b^2];
or (2a-1)^2 + (2b)^2] = (a-2)^2 + b^2;
or (4a^2 - 4a + 1) + 4b^2 = (a^2 - 4a + 4) + b^2;
or 3a^2 + 3b^2 = 3;
or a^2 + b^2 = 1 (a circle centered at (0,0) and radius = 1).
Hence the distance from this circle's center to the point (3,4) = sqrt[(3-0)^2 + (4-0)^2] = sqrt(9 + 16) = 5.