Verified Solution • 2025–26 Session moderate

QJEEM25Ma777b: JEE Main 2025 Mathematics - Class 11 JEEM25Ma777b Solution: Let , be the equation of a circle with center at . If the area…

Let ∣zˉ−i2zˉ+i∣=13,z∈C\left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3}, z \in C , be the equation of a circle with center at CC . If the area of the triangle, whose vertices are at the points (0,0),C(0,0), C and (α,0)(\alpha, 0) is 11 square units, then α2\alpha^2 equals:

Multiple Choice Options

  • A 12125\frac{121}{25}
  • B 100
  • C 8125\frac{81}{25}
  • D 50

Step-by-Step Verified Solution

amp;∣zˉ−i2zˉ+i∣=13amp;∣zˉ−izˉ+i2∣=23amp;3∣x−iy−i∣=2∣x−iy+i2∣amp;9(x2+(y+1)2)=4(x2+(y−1/3)2)amp;9x2+9y2+18y+9=4x2+4y2−4y+1amp;5x2+5y2+22y+8=0amp;x2+y2+225y+85=0amp; centre ⇒(0,−115)\begin{aligned}& \left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3} \\& \left|\frac{\bar{z}-i}{\bar{z}+\frac{i}{2}}\right|=\frac{2}{3} \\& 3|x-i y-i|=2\left|x-i y+\frac{i}{2}\right| \\& 9\left(x^2+(y+1)^2\right)=4\left(x^2+(y-1 / 3)^2\right) \\& 9 x^2+9 y^2+18 y+9=4 x^2+4 y^2-4 y+1 \\& 5 x^2+5 y^2+22 y+8=0 \\& x^2+y^2+\frac{22}{5} y+\frac{8}{5}=0 \\& \text { centre } \Rightarrow\left(0,-\frac{11}{5}\right)\end{aligned}

|12|0010−11/51α01||=11

amp;⇒(−115α)2=(11×2)2amp;⇒α2=100\begin{aligned}& \Rightarrow\left(-\frac{11}{5} \alpha\right)^2=(11 \times 2)^2 \\& \Rightarrow \alpha^2=100\end{aligned}

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