Let the lines (2 – i)z = (2 + i)z̄ and (2 + i)z + (i – 2)z̄ – 4i = 0, (here i² = -1 ) be normal to a

Q: Why do we substitute $z = x + iy$ and $\bar{z} = x - iy$ instead of solving in the complex domain?

While you *can* solve in the complex domain, it often leads to algebraic traps. Converting to Cartesian coordinates maps the problem to a visual space where your intuition for lines and circles is strongest. It is a strategic choice to reduce cognitive load and avoid errors.

Q: Wait, how do I know the lines are not parallel?

That is a great observation! If the coefficients of $x$ and $y$ are proportional, they are parallel. Here, $L_1$ has a slope of $1/2$ and $L_2$ has a slope of $-1/2$. Since the slopes are different, they must intersect. Always check for parallelism before assuming an intersection exists!

Q: Is the radius always positive?

Yes! The distance formula uses an absolute value for a reason. If you get a negative value, you have likely made a sign error in your coordinate substitution. Remember, distance is a physical reality—it must be positive.

Question

Let the lines $(2 – i)z = (2 + i)\bar{z}$ and $(2 + i)z + (i – 2)\bar{z} – 4i = 0$, (here $i^2 = -1$ ) be normal to a circle C. If the line $iz + \bar{z} + 1 + i = 0$ is tangent to this circle C, then its radius is :

Options:

(A) $\frac{3}{\sqrt{2}}$

(B) $3\sqrt{2}$

(C) $\frac{3}{2\sqrt{2}}$

(D) $\frac{1}{2\sqrt{2}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given two lines in the complex plane, $(2 - i)z = (2 + i)\bar{z}$ and $(2 + i)z + (i - 2)\bar{z} - 4i = 0$, which serve as normals to a circle. Since a normal to a circle must pass through its center, the intersection of these two lines provides the center $(h, k)$ of the circle.

To solve this, we employ the **Cartesian bridge** by substituting $z = x + iy$ and $\bar{z} = x - iy$. This translates the complex equations into standard linear equations in $x$ and $y$.

### Translating the Normals

For the first normal, $(2 - i)(x + iy) = (2 + i)(x - iy)$, we expand the terms:
$$2x + 2iy - ix - i^2y = 2x - 2iy + ix + i^2y$$

Recalling that $i^2 = -1$, the equation simplifies significantly:
$$2x + 2iy - ix + y = 2x - 2iy + ix - y$$

Canceling the $2x$ terms and rearranging, we obtain $4iy - 2ix + 2y = 0$. Dividing by $2i$ (or simplifying the real and imaginary components), we arrive at the first line, $L_1$:
$$x - 2y = 0$$

For the second normal, $(2 + i)(x + iy) + (i - 2)(x - iy) - 4i = 0$, we expand:
$$2x + 2iy + ix - y + ix + y - 2x + 2iy - 4i = 0$$

The $2x$ and $-2x$ terms cancel, as do the $-y$ and $+y$ terms. We are left with:
$$4iy + 2ix - 4i = 0$$

Dividing the entire equation by $2i$, we obtain the second line, $L_2$:
$$x + 2y - 2 = 0$$

### Locating the Center

We now solve the system of linear equations:
1) $x - 2y = 0$
2) $x + 2y = 2$

Adding these two equations yields $2x = 2$, which implies $x = 1$. Substituting $x = 1$ into the first equation, we find $1 - 2y = 0$, or $y = 1/2$. Thus, the center of the circle is $(1, 1/2)$.

### Determining the Radius

We are given the tangent line $iz + \bar{z} + 1 + i = 0$. Substituting $z = x + iy$ and $\bar{z} = x - iy$:
$$i(x + iy) + (x - iy) + 1 + i = 0$$
$$ix - y + x - iy + 1 + i = 0$$

Grouping the real and imaginary parts, we get:
$$(x - y + 1) + i(x - y + 1) = 0$$

For this complex equation to hold, both the real and imaginary parts must be zero. This yields the tangent line equation:
$$x - y + 1 = 0$$

The radius $r$ is the perpendicular distance from the center $(1, 1/2)$ to the line $x - y + 1 = 0$. Using the distance formula:
$$r = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$

Substituting our values:
$$r = \frac{|1(1) - 1(1/2) + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|1 - 0.5 + 1|}{\sqrt{2}} = \frac{1.5}{\sqrt{2}}$$

The final radius of the circle is:
**$$r = \frac{3}{2\sqrt{2}}$$**

Animated Solution for Mathematics - Complex Numbers: Let the lines $(2- i) z = (2 + i) \overset{z}{ˉ}$ and $(2 + i) z + (i -2) \overset{z}{ˉ} -4 i = 0$ , (here $i^{2} = - 1$ ) be normal to a circle C. If the line $i z + \overset{z}{ˉ} + 1 + i = 0$ is tangent to this circle C, then its radius is :

Select Answer:

Visualized Solution

Understanding the Geometry

Cartesian Conversion Strategy

Substituting into First Normal

Simplifying First Normal

Substituting into Second Normal

Simplifying Second Normal

Finding the Center of the Circle

Substituting into Tangent Line

Simplifying Tangent Line

Radius as Perpendicular Distance

Applying the Distance Formula

Calculating the Final Radius

Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

Translating the Normals

Locating the Center

Determining the Radius

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