Question

Radius of circle which touches the line iz + $\bar{z}$ + 1 + i = 0 and the lines (2 – i) z = (2 + i) $\bar{z}$ and (2 + i) z + (i – 2) $\bar{z}$ – 4i = 0 are the normal’s to the circles –

• A. $3\sqrt{2}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{3}{2\sqrt{2}}$
• D. None of these

Answer 1

Answer: (C)

$\frac{3}{2\sqrt{2}}$

Answer 2

Sol. Equation of normals of circle are

(2 – i) z = (2 + i) z …(1)

and(2 + i)z + (i – 2)$\bar{z}$ – 4i = 0 …(2)

replace $\bar{z}$ form (2) with the help of (1)

(2 + i)z + $\frac{(i - 2)(2 - i)}{(2 + i)}$z – 4i = 0

Ž (2 + i)z + $\frac{(4i - 3)(2 - i)}{5}$z – 4i = 0

Ž (2 + i)z + $\frac{(11i - 2)}{5}$z = 4i

Ž (16i + 8)z = 20 i

Ž z = $\frac{20i}{16i + 8}$ = $\frac{5i(2 - 4i)}{(2 + 4i)(2 - 4i)}$

centre z = $\left( 1 + \frac{i}{2} \right)$

(Intersection point of two normals) equation of line in standard form

= $\frac{iz}{1 + i}$+$\frac{\bar{z}}{1 + i}$ + 1 = 0

(1 + i) z + (1 – i)$\bar{z}$ + 2 = 0 radius of circle

= $\frac{\left| \frac{(1 + i)(2 + i)}{2} + \frac{(1 - i)(2 - i)}{2} + 2 \right|}{|1 + i| + |1 - i|}$ = $\frac{3}{2\sqrt{2}}$.