JEE PYQ: Complex Numbers Question 11
Question 11 - 2021 (25 Feb Shift 1)
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:
(1) $\frac{3}{\sqrt{2}}$
(2) $3\sqrt{2}$
(3) $\frac{3}{2\sqrt{2}}$
(4) $\frac{1}{2\sqrt{2}}$
Show Answer
Answer: (3)
Solution
First normal: $(2-i)(x+iy) = (2+i)(x-iy)$ gives $y = 0$. Second normal: simplifies to $x = 1$. Center $(1, 0)$. Tangent: $iz + \bar{z} + 1 + i = 0$ becomes $-y + ix + x - iy + 1 + i = 0$, giving $x - y + 1 = 0$ and $x - y + 1 = 0$. Distance from $(1,0)$: $\frac{|1+1|}{\sqrt{2}} = \frac{3}{2\sqrt{2}}$.
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