JEE MAIN - Mathematics (2021 - 25th February Morning Shift - No. 8)
Let the lines (2 $$-$$ i)z = (2 + i)$$\overline z $$ and (2 $$+$$ i)z + (i $$-$$ 2)$$\overline z $$ $$-$$ 4i = 0, (here i2 = $$-$$1) be normal to a circle C. If the line iz + $$\overline z $$ + 1 + i = 0 is tangent to this circle C, then its radius is :
$${3 \over {2\sqrt 2 }}$$
$$3\sqrt 2 $$
$${1 \over {2\sqrt 2 }}$$
$${3 \over {\sqrt 2 }}$$
Explanation
$$(2 - i)z = (2 + i)\overline z $$
$$ \Rightarrow (2 - i)(x + iy) = (2 + i)(x - iy)$$
$$ \Rightarrow 2x - ix + 2iy + y = 2x + ix - 2 - iy + y$$
$$ \Rightarrow 2ix - 4iy = 0$$
$${L_1}:x - 2y = 0$$
$$ \Rightarrow (2 + i)z + (i - 2)\overline z - 4i = 0$$
$$ \Rightarrow (2 + i)(x + iy) + (i - 2)(x - iy) - 4i = 0$$
$$ \Rightarrow 2x + ix + 2iy - y + ix - 2x + y + 2iy - 4i = 0$$
$$ \Rightarrow 2ix + 4iy - 4i = 0$$
$${L_2}:x + 2y - 2 = 0$$
Solve L1 and L2: x = 1, $$4y = 2,y = {1 \over 2}$$
$$ \therefore $$ $$x = 1$$
Centre $$\left( {1,{1 \over 2}} \right)$$
$${L_3}:iz + \overline z + 1 + i = 0$$
$$ \Rightarrow i(x + iy) + x - iy + 1 + i = 0$$
$$ \Rightarrow ix - y + x - iy + 1 + i = 0$$
$$ \Rightarrow (x - y + 1) + i(x - y + 1) = 0$$
Radius = distance from $$\left( {1,{1 \over 2}} \right)$$ to $$x - y + 1 = 0$$
$$ \Rightarrow $$ $$r = {{1 - {1 \over 2} + 1} \over {\sqrt 2 }}$$
$$ \Rightarrow $$ $$r = {3 \over {2\sqrt 2 }}$$
$$ \Rightarrow (2 - i)(x + iy) = (2 + i)(x - iy)$$
$$ \Rightarrow 2x - ix + 2iy + y = 2x + ix - 2 - iy + y$$
$$ \Rightarrow 2ix - 4iy = 0$$
$${L_1}:x - 2y = 0$$
$$ \Rightarrow (2 + i)z + (i - 2)\overline z - 4i = 0$$
$$ \Rightarrow (2 + i)(x + iy) + (i - 2)(x - iy) - 4i = 0$$
$$ \Rightarrow 2x + ix + 2iy - y + ix - 2x + y + 2iy - 4i = 0$$
$$ \Rightarrow 2ix + 4iy - 4i = 0$$
$${L_2}:x + 2y - 2 = 0$$
Solve L1 and L2: x = 1, $$4y = 2,y = {1 \over 2}$$
$$ \therefore $$ $$x = 1$$
Centre $$\left( {1,{1 \over 2}} \right)$$
$${L_3}:iz + \overline z + 1 + i = 0$$
$$ \Rightarrow i(x + iy) + x - iy + 1 + i = 0$$
$$ \Rightarrow ix - y + x - iy + 1 + i = 0$$
$$ \Rightarrow (x - y + 1) + i(x - y + 1) = 0$$
Radius = distance from $$\left( {1,{1 \over 2}} \right)$$ to $$x - y + 1 = 0$$
$$ \Rightarrow $$ $$r = {{1 - {1 \over 2} + 1} \over {\sqrt 2 }}$$
$$ \Rightarrow $$ $$r = {3 \over {2\sqrt 2 }}$$
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