Question:
If \(z_1=8+4i,\; z_2=6+4i\) and
\[
\operatorname{Arg}\left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{4},
\]
then \(z\) satisfies
If \(z_1=8+4i,\; z_2=6+4i\) and
\[
\operatorname{Arg}\left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{4},
\]
then \(z\) satisfies
Show Hint
The locus
\[
\operatorname{Arg}\left(\frac{z-z_1}{z-z_2}\right)=\alpha
\]
represents the circle through \(z_1\) and \(z_2\) from which the chord \(z_1z_2\) subtends a constant angle \(\alpha\). Use
\[
AB=2R\sin\alpha
\]
to find the radius and then determine the centre from the perpendicular bisector of the chord.
Updated On: Jul 9, 2026
- \( |z-7-4i|=1 \)
- \( |z-7-5i|=\sqrt{2} \)
- \( |z-4i|=8 \)
- \( |z-1-7i|=\sqrt{18} \) \bigskip
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
If
\[
\operatorname{Arg}\left(\frac{z-z_1}{z-z_2}\right)=\alpha,
\]
then the angle subtended by the line segment joining \(z_1\) and \(z_2\) at the point \(z\) is \(\alpha\).
Thus, the locus of \(z\) is a circle passing through \(z_1\) and \(z_2\) such that the chord \(z_1z_2\) subtends a constant angle \(\alpha\) at every point on the circle.
Step 1: Identify the points represented by \(z_1\) and \(z_2\). Given \[ z_1=8+4i,\qquad z_2=6+4i. \] Therefore, \[ A(8,4),\qquad B(6,4). \] The length of the chord \(AB\) is \[ AB=\sqrt{(8-6)^2+(4-4)^2}=2. \]
Step 2: Use the condition that the chord subtends an angle \(\frac{\pi}{4}\). For a circle of radius \(R\), \[ AB=2R\sin\theta, \] where \(\theta\) is the angle subtended by the chord at a point on the circumference. Here, \[ AB=2, \qquad \theta=\frac{\pi}{4}. \] Hence, \[ 2=2R\sin\frac{\pi}{4}. \] \[ 2=2R\left(\frac{\sqrt2}{2}\right). \] \[ R=\sqrt2. \]
Step 3: Find the centre of the circle. The midpoint of \(AB\) is \[ M\left(\frac{8+6}{2},\frac{4+4}{2}\right) =(7,4). \] Since the chord is horizontal, the centre lies on the perpendicular bisector \[ x=7. \] Let the centre be \[ C=(7,k). \] Using \[ CM^2=R^2-\left(\frac{AB}{2}\right)^2, \] we get \[ CM^2=(\sqrt2)^2-1^2 =2-1=1. \] Hence, \[ CM=1. \] Therefore, \[ k=4\pm1. \] So the possible centres are \[ (7,5)\quad \text{and}\quad (7,3). \]
Step 4: Determine the correct centre using the given argument. The condition \[ \operatorname{Arg}\left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{4} \] corresponds to the major arc lying above the chord \(AB\). Hence the required circle has centre \[ (7,5). \] Its radius is \[ \sqrt2. \] Therefore, the equation of the locus is \[ (x-7)^2+(y-5)^2=2. \] In complex form, \[ |z-(7+5i)|=\sqrt2. \] \[ |z-7-5i|=\sqrt2. \]
Step 5: Write the final answer. \[ \boxed{|z-7-5i|=\sqrt2} \]
Step 1: Identify the points represented by \(z_1\) and \(z_2\). Given \[ z_1=8+4i,\qquad z_2=6+4i. \] Therefore, \[ A(8,4),\qquad B(6,4). \] The length of the chord \(AB\) is \[ AB=\sqrt{(8-6)^2+(4-4)^2}=2. \]
Step 2: Use the condition that the chord subtends an angle \(\frac{\pi}{4}\). For a circle of radius \(R\), \[ AB=2R\sin\theta, \] where \(\theta\) is the angle subtended by the chord at a point on the circumference. Here, \[ AB=2, \qquad \theta=\frac{\pi}{4}. \] Hence, \[ 2=2R\sin\frac{\pi}{4}. \] \[ 2=2R\left(\frac{\sqrt2}{2}\right). \] \[ R=\sqrt2. \]
Step 3: Find the centre of the circle. The midpoint of \(AB\) is \[ M\left(\frac{8+6}{2},\frac{4+4}{2}\right) =(7,4). \] Since the chord is horizontal, the centre lies on the perpendicular bisector \[ x=7. \] Let the centre be \[ C=(7,k). \] Using \[ CM^2=R^2-\left(\frac{AB}{2}\right)^2, \] we get \[ CM^2=(\sqrt2)^2-1^2 =2-1=1. \] Hence, \[ CM=1. \] Therefore, \[ k=4\pm1. \] So the possible centres are \[ (7,5)\quad \text{and}\quad (7,3). \]
Step 4: Determine the correct centre using the given argument. The condition \[ \operatorname{Arg}\left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{4} \] corresponds to the major arc lying above the chord \(AB\). Hence the required circle has centre \[ (7,5). \] Its radius is \[ \sqrt2. \] Therefore, the equation of the locus is \[ (x-7)^2+(y-5)^2=2. \] In complex form, \[ |z-(7+5i)|=\sqrt2. \] \[ |z-7-5i|=\sqrt2. \]
Step 5: Write the final answer. \[ \boxed{|z-7-5i|=\sqrt2} \]
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