© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Express the given lines in terms of x and y
We interpret the complex number
$z = x + i\,y$
and its conjugate
$\overline{z} = x - i\,y.$
The first given line is:
$$(2 - i)z = (2 + i)\overline{z}.$$
Substitute
$z = x + i\,y$
and
$\overline{z} = x - i\,y$
into this equation:
$$(2 - i)(x + i\,y) = (2 + i)(x - i\,y).$$
Step 2: Simplify the first line
Expand both sides:
Left side:
$$(2 - i)(x + i\,y) = 2x + 2i\,y - i\,x - i^2\,y = 2x + 2i\,y - i\,x + y,$$
because
$i^2 = -1.$
Right side:
$$(2 + i)(x - i\,y) = 2x - 2i^2\,y + i\,x - i^2\,y = 2x + i\,x + 2y - i\,y,$$
but carefully expanding, we have
$$ (2 + i)(x - i\,y) = 2x - 2i\,y + i\,x - i^2\,y = 2x - 2i\,y + i\,x + y.$$
Simplifying each side carefully gives:
Left side = $2x - i\,x + 2i\,y + y,$
Right side = $2x + i\,x - 2i\,y + y.$
Setting them equal:
$$2x - i\,x + 2i\,y + y = 2x + i\,x - 2i\,y + y.$$
Rearrange terms to isolate those involving $i$:
$$( - i\,x + 2i\,y ) - ( i\,x + (-2i\,y) ) = 0.$$
This simplifies to:
$$- i\,x + 2i\,y - i\,x + 2i\,y = 0 \quad \Longrightarrow \quad -2i\,x + 4i\,y = 0 \quad \Longrightarrow \quad 2i(x - 2y) = 0.$$
Dividing out the non-zero constant,
$$x - 2y = 0.$$
So the first line
$L_1$
is:
$$L_1: \, x - 2y = 0.$$
Step 3: Simplify the second line
The second given line is:
$$(2 + i)z + (i - 2)\overline{z} - 4i = 0.$$
Substitute
$z = x + i\,y$
and
$\overline{z} = x - i\,y$:
$$(2 + i)(x + i\,y) + (i - 2)(x - i\,y) - 4i = 0.$$
Step 4: Expand and combine like terms
Expand each term:
$(2 + i)(x + i\,y) = 2x + 2i\,y + i\,x + i^2\,y = 2x + 2i\,y + i\,x - y,$
$(i - 2)(x - i\,y) = i\,x - i^2\,y - 2x + 2i\,y = i\,x + y - 2x + 2i\,y.$
Summing these and subtracting $4i$:
\[
(2x + i\,x - y) + (2i\,y + 2i\,y) + (i\,x + y - 2x + 2i\,y) - 4i = 0.
\]
Combine like terms carefully:
• Real parts: $(2x - 2x) + ( -y + y ) = 0,$
• Terms with $x$: $i\,x + i\,x = 2i\,x,$
• Terms with $y$: $2i\,y + 2i\,y + 2i\,y = 6i\,y,$
• Also subtract $4i.$
Let us group them to see the net effect:
\[
2i\,x + 6i\,y - 4i = i \bigl(2x + 6y - 4 \bigr) = 0.
\]
Hence,
\[
2x + 6y - 4 = 0 \quad \Longrightarrow \quad x + 3y - 2 = 0.
\]
However, the solution provided in the reference simplified to
$x + 2y - 2 = 0.$
Let us recheck the combination carefully:
Indeed, in many standard solutions, after correct simplification, we get:
$(2 + i)(x + i\,y) + (i - 2)(x - i\,y) - 4i = 0$
→ $2x + ix + 2i\,y - y + (ix + y - 2x + 2i\,y) - 4i = 0.$
Summing:
$2x - 2x + ix + ix + 2i\,y + 2i\,y + y - y - 4i.$
= $2ix + 4i\,y - 4i.$
Factor out $2i$:
$2i(x + 2y - 2) = 0.$
Thus
$$x + 2y - 2 = 0.$$
So the second line
$L_2$
is:
$$L_2: \, x + 2y - 2 = 0.$$
Step 5: Find the intersection of the two lines (center of the circle)
The lines
$L_1: x - 2y = 0 \quad (1)$
and
$L_2: x + 2y - 2 = 0 \quad (2)$
represent normals to the circle. Their intersection gives the center of the circle. Solve simultaneously:
From $L_1,$ $x = 2y.$
Substitute into $L_2$:
$$(2y) + 2y - 2 = 0 \quad \Longrightarrow \quad 4y - 2 = 0 \quad \Longrightarrow \quad 4y = 2 \quad \Longrightarrow \quad y = \tfrac12.$$
Hence, $x = 2 \times \tfrac12 = 1.$
So the center is
$$(1, \tfrac12).$$
Step 6: Convert the tangent line to Cartesian form
The line given for tangency is:
$$i\,z + \overline{z} + 1 + i = 0.$$
Substitute $z = x + i\,y$ and $\overline{z} = x - i\,y,$
$$
i(x + i\,y) + (x - i\,y) + 1 + i = 0.
$$
Expand:
$$
\bigl(i\,x - y\bigr) + \bigl(x - i\,y\bigr) + 1 + i = 0.
$$
Combine like terms:
• Real part: $(x - y + 1)$
• Imag part: $(i\,x - i\,y + i)$ → factor out $i$ → $i(x - y + 1)$
So the sum is
$$(x - y + 1) + i(x - y + 1).$$
For this to be zero, both real and imaginary parts must be zero, which yields:
$$x - y + 1 = 0.$$
Thus, in Cartesian form, the tangent line is:
$$x - y + 1 = 0.$$
Step 7: Compute the circle's radius using perpendicular distance
The radius $r$ of the circle is the perpendicular distance from its center
$(1, \tfrac12)$
to the tangent line
$x - y + 1 = 0.$
The formula for perpendicular distance from a point $(x_0, y_0)$ to the line
$ax + b\,y + c = 0$
is
$$
\text{Distance}
= \frac{\bigl|\,a\,x_0 + b\,y_0 + c\,\bigr|}{\sqrt{a^2 + b^2}}.
$$
Here, $a = 1,\; b = -1,\; c = 1,\; x_0 = 1,\; y_0 = \tfrac12.$
So,
\[
r
= \frac{\bigl|1\cdot 1 + (-1)\cdot \tfrac12 + 1 \bigr|}{\sqrt{1^2 + (-1)^2}}
= \frac{\left|1 - \tfrac12 + 1 \right|}{\sqrt{1 + 1}}
= \frac{\left|\tfrac32\right|}{\sqrt{2}}
= \frac{\tfrac32}{\sqrt{2}}
= \frac{3}{2\sqrt{2}}.
\]
Step 8: Conclude the final answer
Hence, the radius of the circle $C$ is:
$$
\boxed{\frac{3}{2\sqrt{2}}}.
$$
