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Complex Numbers Concyclic Points with Origin | 2/z1 = 1/z2 + 1/z3

JEE Maths question with a full step-by-step solution.

Question
If z1,z2,z3z_1, z_2, z_3 are the non-zero complex numbers representing the points A,B,CA, B, C such that
2z1=1z2+1z3,\frac{2}{z_1} = \frac{1}{z_2} + \frac{1}{z_3},
 then which of the following is true?
AA,B,CA, B, C are collinear.
BA circle passes through the points A,B,CA, B, C and has centre at the origin OO.
CA circle passes through the points A,B,CA, B, C and passes through the origin.correct
DNone of these.
Solution
Step 1: Rearrange the given condition. Starting from
2z1=1z2+1z3,\frac{2}{z_1} = \frac{1}{z_2} + \frac{1}{z_3},
 split the left side as 1z1+1z1\dfrac{1}{z_1} + \dfrac{1}{z_1} and regroup:
1z1−1z2=1z3−1z1.\frac{1}{z_1} - \frac{1}{z_2} = \frac{1}{z_3} - \frac{1}{z_1}.
 Step 2: Combine each side into a single fraction.
z2−z1z1z2=z1−z3z1z3.\frac{z_2 - z_1}{z_1 z_2} = \frac{z_1 - z_3}{z_1 z_3}.
 Since z1≠0z_1 \neq 0, cancel z1z_1 from both denominators
z2−z1z2=z1−z3z3.\frac{z_2 - z_1}{z_2} = \frac{z_1 - z_3}{z_3}.
 Step 3: Solve for the ratio of differences. Cross-multiplying gives
z2−z1z1−z3=z2z3,\frac{z_2 - z_1}{z_1 - z_3} = \frac{z_2}{z_3},
 and replacing z1−z3=−(z3−z1)z_1 - z_3 = -(z_3 - z_1),
z2−z1z3−z1=−z2z3.\frac{z_2 - z_1}{z_3 - z_1} = -\frac{z_2}{z_3}.
 Step 4: Take the argument of both sides.
Arg⁡(z2−z1z3−z1)=Arg⁡(−z2z3)=Arg⁡(z2z3)±π.\operatorname{Arg}\left(\frac{z_2 - z_1}{z_3 - z_1}\right) = \operatorname{Arg}\left(-\frac{z_2}{z_3}\right) = \operatorname{Arg}\left(\frac{z_2}{z_3}\right) \pm \pi.
 Step 5: Express the right side relative to the origin. Since z2z3=z2−0z3−0\dfrac{z_2}{z_3} = \dfrac{z_2 - 0}{z_3 - 0},
Arg⁡(z2−z1z3−z1)−Arg⁡(z2−0z3−0)=±π.\operatorname{Arg}\left(\frac{z_2 - z_1}{z_3 - z_1}\right) - \operatorname{Arg}\left(\frac{z_2 - 0}{z_3 - 0}\right) = \pm\pi.
 Step 6: Interpret geometrically. The term Arg⁡(z2−z1z3−z1)\operatorname{Arg}\left(\dfrac{z_2 - z_1}{z_3 - z_1}\right) is the angle subtended by the segment BCBC at the point A(z1)A(z_1), and Arg⁡(z2−0z3−0)\operatorname{Arg}\left(\dfrac{z_2 - 0}{z_3 - 0}\right) is the angle subtended by BCBC at the origin OO. Their difference equals ±π\pm\pi, which means the inscribed angles that BCBC subtends at AA and at OO are supplementary. Consequently AA and OO lie on opposite arcs of the circle determined by BB and CC, so the four points O,A,B,CO, A, B, C are concyclic. Therefore a circle passes through the points A,B,CA, B, C and also passes through the origin. (See the figure: the chord BCBC subtends supplementary angles at AA and at OO, so O,A,B,CO, A, B, C lie on one circle.) Correct answer: (3)
Solution working
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