Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Complex Numbers question

2021 · 25 Feb · Shift 1 · Q31
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Complex Numbers
  5. /2021 · 25 Feb · Shift 1 · Q31

Complex Numbers question

2021 · 25 Feb · Shift 1 · Q31

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let the lines (2 −-− i)z = (2 + i) z‾\overline zz and (2 +++ i)z + (i −-− 2) z‾−\overline z -z− 4i = 0, (here i2 =−-− 1) be normal to a circle C. If the line iz + z‾\overline zz + 1 + i = 0 is tangent to this circle C, then its radius is :
  1. A
    322{3 \over {2\sqrt 2 }}22​3​
  2. B
    323\sqrt 232​
  3. C
    122{1 \over {2\sqrt 2 }}22​1​
  4. D
    32{3 \over {\sqrt 2 }}2​3​
View written solutionFree

Correct answer: A

Let z=x+iyz=x+iyz=x+iy so that z‾=x−iy\overline z=x-iyz=x−iy.

We are given two normals to a circle CCC, hence both pass through the centre of the circle. So the centre is the intersection point of these two lines.

1. Convert the first line to Cartesian form

Given

(2−i)z=(2+i)z‾.(2-i)z=(2+i)\overline z.(2−i)z=(2+i)z.

Substitute z=x+iyz=x+iyz=x+iy, z‾=x−iy\overline z=x-iyz=x−iy:

(2−i)(x+iy)=(2+i)(x−iy).(2-i)(x+iy)=(2+i)(x-iy).(2−i)(x+iy)=(2+i)(x−iy).

Expand both sides:

Left side:

(2−i)(x+iy)=2x+2iy−ix−i2y=2x+y+i(2y−x).(2-i)(x+iy)=2x+2iy-ix-i^2y=2x+y+i(2y-x).(2−i)(x+iy)=2x+2iy−ix−i2y=2x+y+i(2y−x).

Right side:

(2+i)(x−iy)=2x−2iy+ix−i2y=2x+y+i(x−2y).(2+i)(x-iy)=2x-2iy+ix-i^2y=2x+y+i(x-2y).(2+i)(x−iy)=2x−2iy+ix−i2y=2x+y+i(x−2y).

Equating imaginary parts:

2y−x=x−2y2y-x=x-2y2y−x=x−2y 4y=2x4y=2x4y=2x x=2y.x=2y.x=2y.

So first line is

x−2y=0.x-2y=0.x−2y=0.

2. Convert the second line to Cartesian form

Given

(2+i)z+(i−2)z‾−4i=0.(2+i)z+(i-2)\overline z-4i=0.(2+i)z+(i−2)z−4i=0.

Now,

(2+i)z=(2+i)(x+iy)=2x−y+i(x+2y),(2+i)z=(2+i)(x+iy)=2x-y+i(x+2y),(2+i)z=(2+i)(x+iy)=2x−y+i(x+2y), (i−2)z‾=(i−2)(x−iy)=−2x−y+i(x+2y).(i-2)\overline z=(i-2)(x-iy)=-2x-y+i(x+2y).(i−2)z=(i−2)(x−iy)=−2x−y+i(x+2y).

Adding,

(2+i)z+(i−2)z‾=(−2y)+i(2x+4y).(2+i)z+(i-2)\overline z = (-2y)+i(2x+4y).(2+i)z+(i−2)z=(−2y)+i(2x+4y).

So the equation becomes

−2y+i(2x+4y)−4i=0.-2y+i(2x+4y)-4i=0.−2y+i(2x+4y)−4i=0.

For this complex number to be zero, real and imaginary parts must both vanish:

−2y=0⇒y=0,-2y=0 \Rightarrow y=0,−2y=0⇒y=0, 2x+4y−4=0⇒2x−4=0⇒x=2.2x+4y-4=0 \Rightarrow 2x-4=0 \Rightarrow x=2.2x+4y−4=0⇒2x−4=0⇒x=2.

Thus this represents the point (2,0)(2,0)(2,0).

Since it is said to be a normal to the circle, this must mean the normal line passes through this point and the centre; combined with the first normal, the centre is obtained as the intersection consistent with the standard line form. A better way is to rewrite the second equation directly in line form using the standard conversion.

Let us instead use the standard identities:

z+z‾=2x,z−z‾=2iy.z+\overline z=2x,\qquad z-\overline z=2iy.z+z=2x,z−z=2iy.

Given

(2+i)z+(i−2)z‾−4i=0.(2+i)z+(i-2)\overline z-4i=0.(2+i)z+(i−2)z−4i=0.

Group terms:

2(z−z‾)+i(z+z‾)−4i=0.2(z-\overline z)+i(z+\overline z)-4i=0.2(z−z)+i(z+z)−4i=0.

Now substitute:

2(2iy)+i(2x)−4i=02(2iy)+i(2x)-4i=02(2iy)+i(2x)−4i=0 4iy+2ix−4i=04iy+2ix-4i=04iy+2ix−4i=0 i(2x+4y−4)=0.i(2x+4y-4)=0.i(2x+4y−4)=0.

Hence

2x+4y−4=02x+4y-4=02x+4y−4=0 x+2y−2=0.x+2y-2=0.x+2y−2=0.

So the second normal is the line

x+2y−2=0.x+2y-2=0.x+2y−2=0.

3. Find the centre of the circle

The centre is the intersection of

x−2y=0,x-2y=0,x−2y=0, x+2y−2=0.x+2y-2=0.x+2y−2=0.

From the first,

x=2y.x=2y.x=2y.

Substitute into the second:

2y+2y−2=02y+2y-2=02y+2y−2=0 4y=24y=24y=2 y=12.y=\frac12.y=21​.

Then

x=1.x=1.x=1.

So centre is

(1,12).(1,\tfrac12).(1,21​).

4. Convert the tangent line to Cartesian form

Given tangent:

iz+z‾+1+i=0.iz+\overline z+1+i=0.iz+z+1+i=0.

Substitute z=x+iyz=x+iyz=x+iy, z‾=x−iy\overline z=x-iyz=x−iy. First,

iz=i(x+iy)=ix−y.iz=i(x+iy)=ix-y.iz=i(x+iy)=ix−y.

So

iz+z‾+1+i=(ix−y)+(x−iy)+1+i.iz+\overline z+1+i = (ix-y)+(x-iy)+1+i.iz+z+1+i=(ix−y)+(x−iy)+1+i.

Real part:

x−y+1,x-y+1,x−y+1,

Imaginary part:

x−y+1.x-y+1.x−y+1.

Thus

(1+i)(x−y+1)=0(1+i)(x-y+1)=0(1+i)(x−y+1)=0

Rightarrow x-y+1=0.

SotangentlineisSo tangent line isSotangentlineis

x-y+1=0.

--- ## 5. Radius = distance from centre to tangent Distance from point $(x_0,y_0)=(1,\tfrac12)$ to line

x-y+1=0

isisis

r=\frac{|1-\tfrac12+1|}{\sqrt{1^2+(-1)^2}} =\frac{|\tfrac32|}{\sqrt2} =\frac{3}{2\sqrt2}.

Sotheradiusis So the radius isSotheradiusis

\boxed{\frac{3}{2\sqrt2}}.

--- ## 6. Check with options This matches **Option A**. --- ## 7. Compare with stored correct answer Stored correct answer: **A** Our derived answer: **A** So they agree.
PreviousNext

More from Complex Numbers

  • If α, β∈ R are such that 1 − 2i (here i2 =− 1) is a root of z2 + α z + β = 0, then (α−β) is equal to :2021 · MCQ
  • Let S={n∈N​(01​i0​)n(ac​bd​)=(ac​bd​)∀a,b,c,d∈R}…2021 · Numerical
  • The equation of a circle is Re(z2) + 2(Im(z))2 + 2Re(z) = 0, where z = x + iy. A line which passes through the center of the given circle and the vertex of the parabola, x2 − 6x − y + 13 = 0, has y-intercept equal to ​…2021 · Numerical
  • The equation arg(z+1z−1​)=4π​ represents a circle with :2021 · MCQ
  • Let z=21−i3​​, i=−1​. Then the value of 21+(z+z1​)3+(z2+z21​)3+(z3+z31​)3+....+(z21+z211​)3…2021 · Numerical
  • If (3​+i)100=299(p+iq), then p and q are roots of the equation :2021 · MCQ
  • The least positive integer n such that (1−i)n−2(2i)n​,i=−1​ is a positive integer, is ​.2021 · Numerical
  • Let z be those complex numbers which satisfy | z + 5 | ≤ 4 and z(1 + i) + z(1 − i) ≥− 10, i = −1​. If the maximum value of | z + 1 |2 is α+β2​, then the value of (α+β) is ​…2021 · Numerical