Verified Solution • 2025–26 Session moderate

QJEEM17Mae7fb: JEE Main 2017 Mathematics - Class 11 JEEM17Mae7fb Solution: Let z C, the set of complex numbers. Then the equation, 2

Let z ∈\in C, the set of complex numbers. Then the equation, 2|z + 3i| −- |z −- i| = 0 represents :

Multiple Choice Options

  • A a circle with radius 83.{8 \over 3}.
  • B a circle with diameter 103.{{10} \over 3}.
  • C an ellipse with length of major axis 163.{{16} \over 3}.
  • D an ellipse with length of minor axis 169.{{16} \over 9}.

Step-by-Step Verified Solution

Given,

2  ∣ \,\left| \, \right. z + 3i  ∣ \,\left| \, \right. =  ∣ \,\left| \, \right. z −- i  ∣ \,\left| \, \right.

Let z = x + iy

⇒\Rightarrow    \,\,\, 2  ∣ \,\left| \, \right. x + iy + 3i  ∣ \,\left| \, \right. =  ∣ \,\left| \, \right. x + iy −- i  ∣ \,\left| \, \right.

⇒\Rightarrow    \,\,\, 2  ∣ \,\left| \, \right. x + i (y + 3)  ∣ \,\left| \, \right. =  ∣ \,\left| \, \right. x + i (y −- 1)  ∣ \,\left| \, \right.

⇒\Rightarrow    \,\,\, 2 x2+(y+3)2\sqrt {{x^2} + {{\left( {y + 3} \right)}^2}} = x2+(y−1)2\sqrt {{x^2} + {{\left( {y - 1} \right)}^2}}

⇒\Rightarrow    \,\,\, 4 (x2 + y2 + 6y + 9) = x2 + y2 −- 2y + 1

⇒\Rightarrow    \,\,\, 3x2 + 3y2 + 26y + 35 = 0

⇒\Rightarrow    \,\,\, x2 + y2 + 263{{26} \over 3} y + 353{{35} \over 3} = 0

This is a equation of circle with center ( −- 133{{13} \over 3} , 0)

∴   \therefore\,\,\, Radius = 0+1699−353\sqrt {0 + {{169} \over 9} - {{35} \over 3}}

= 649\sqrt {{{64} \over 9}}

= 83{8 \over 3}

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