Intercept made by the circle `zbarz+bara+a barz+r=0` on the real axis on complex plane is

Options

A. `(q)/(rcdotn)`
B. `(icdotn)/(q)`
C. `(rcdotn)q`
D. `(q)/(|n|)`

Answer

To find the intercept made by the circle defined by the equation \( z \bar{z} + \alpha \bar{z} + \bar{\alpha} z + R = 0 \) on the real axis in the complex plane, we can follow these steps:

### Step 1: Rewrite the equation
Given the equation:
\[ z \bar{z} + \alpha \bar{z} + \bar{\alpha} z + R = 0 \]
Since we are interested in the intercept on the real axis, we can set \( z = \bar{z} \). Thus, we can rewrite the equation as:
\[ z^2 + \alpha z + \bar{\alpha} z + R = 0 \]

### Step 2: Combine like terms
Combine the terms involving \( z \):
\[ z^2 + (\alpha + \bar{\alpha}) z + R = 0 \]

### Step 3: Identify coefficients
This is a quadratic equation of the form \( az^2 + bz + c = 0 \), where:
- \( a = 1 \)
- \( b = \alpha + \bar{\alpha} \)
- \( c = R \)

### Step 4: Use the quadratic formula
The roots of the quadratic equation can be found using the quadratic formula:
\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ z = \frac{-(\alpha + \bar{\alpha}) \pm \sqrt{(\alpha + \bar{\alpha})^2 - 4R}}{2} \]

### Step 5: Find the intercept
The intercept on the real axis is given by the distance between the two roots \( z_1 \) and \( z_2 \):
\[ |z_1 - z_2| = \left| \frac{\sqrt{(\alpha + \bar{\alpha})^2 - 4R}}{1} \right| = \sqrt{(\alpha + \bar{\alpha})^2 - 4R} \]
Thus, the intercept made by the circle on the real axis is:
\[ \text{Intercept} = \sqrt{(\alpha + \bar{\alpha})^2 - 4R} \]

### Final Answer
The intercept made by the circle on the real axis is:
\[ \sqrt{(\alpha + \bar{\alpha})^2 - 4R} \]

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