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JEE Mathematics 2025 Question with Solution

Let the curve z(1+i)+z‾(1−i)=4, z∈Cz(1 + i) + \overline{z}(1 - i) = 4, \, z \in \mathbb{C}, divide the region ∣z−3∣≤1|z - 3| \leq 1 into two parts of areas α\alpha and β\beta. Then ∣α−β∣|\alpha - \beta| equals:

  • A

    1+π41 + \frac{\pi}{4}

  • B

    1+π21 + \frac{\pi}{2}

  • C

    1+π31 + \frac{\pi}{3}

  • D

    1+π61 + \frac{\pi}{6}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: z(1+i)+z‾(1−i)=4z(1+i)+\overline{z}(1-i)=4 and the region ∣z−3∣≤1|z-3|\le 1.

Find: ∣α−β∣|\alpha-\beta|, where the given line divides the circular region into two parts.

Let z=x+iyz=x+iy, so z‾=x−iy\overline{z}=x-iy. Then

z(1+i)=(x+iy)(1+i)=(x−y)+i(x+y)z(1+i)=(x+iy)(1+i)=(x-y)+i(x+y)

and

z‾(1−i)=(x−iy)(1−i)=(x−y)−i(x+y)\overline{z}(1-i)=(x-iy)(1-i)=(x-y)-i(x+y)

Adding,

z(1+i)+z‾(1−i)=2(x−y)z(1+i)+\overline{z}(1-i)=2(x-y)

Hence,

2(x−y)=42(x-y)=4

so the curve is the straight line

x−y=2x-y=2

or y=x−2y=x-2.

Now,

∣z−3∣≤1|z-3|\le 1

represents the circle with center C(3,0)C(3,0) and radius r=1r=1.

The perpendicular distance of the center C(3,0)C(3,0) from the line x−y−2=0x-y-2=0 is

d=∣3−0−2∣12+(−1)2=12d=\frac{|3-0-2|}{\sqrt{1^2+(-1)^2}}=\frac{1}{\sqrt{2}}

Since $$d

For a circle of radius rr, if a chord is at distance dd from the center, then the area of the smaller segment is

Asmall=r2cos⁡−1(dr)−dr2−d2A_{\text{small}}=r^2\cos^{-1}\left(\frac{d}{r}\right)-d\sqrt{r^2-d^2}

Substituting r=1r=1 and d=12d=\frac{1}{\sqrt{2}},

Asmall=cos⁡−1(12)−12⋅1−12A_{\text{small}}=\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)-\frac{1}{\sqrt{2}}\cdot\sqrt{1-\frac{1}{2}} Asmall=π4−12A_{\text{small}}=\frac{\pi}{4}-\frac{1}{2}

Area Difference Computation

The total area of the circle is

πr2=π\pi r^2=\pi

If the smaller part has area AsmallA_{\text{small}}, then the larger part has area

π−Asmall\pi-A_{\text{small}}

Therefore,

∣α−β∣=(π−Asmall)−Asmall=π−2Asmall|\alpha-\beta|=(\pi-A_{\text{small}})-A_{\text{small}}=\pi-2A_{\text{small}}

Common mistakes

  • Using the incorrect expansion of z‾(1−i)\overline{z}(1-i). This gives the wrong line equation. Write z=x+iyz=x+iy and z‾=x−iy\overline{z}=x-iy carefully, then expand both terms before adding.

  • Assuming the line passes through the center of the circle. The center is (3,0)(3,0), but substituting into x−y−2=0x-y-2=0 gives 1≠01\ne 0, so the two parts are unequal.

  • Using the sector area directly as the segment area. A segment area equals sector area minus triangle area. Use Asmall=r2cos⁡−1(d/r)−dr2−d2A_{\text{small}}=r^2\cos^{-1}(d/r)-d\sqrt{r^2-d^2} instead.

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