Verified Solution • 2025–26 Session moderate

QJEEM21Madc8f: JEE Main 2021 Mathematics - Class 11 JEEM21Madc8f Solution: If the locus of the mid-point of the line segment from the…

If the locus of the mid-point of the line segment from the point (3, 2) to a point on the circle, x2 + y2 = 1 is a circle of radius r, then r is equal to :

Multiple Choice Options

  • A 14{1 \over 4}
  • B 12{1 \over 2}
  • C 1
  • D 13{1 \over 3}

Step-by-Step Verified Solution

Let P(h, k) and point on the circle is (cos θ\theta , sin θ\theta )

∴\therefore 3+cos⁡θ2=h{{3 + \cos \theta } \over 2} = h and 2+sin⁡θ2=k{{2 + \sin \theta } \over 2} = k

cos θ\theta = 2h −- 3 and sin θ\theta = 2h −- 2

Squaring and adding we get

(2h−3)2+(2h−2)2=1{(2h - 3)^2} + {(2h - 2)^2} = 1

⇒4x2−12x+9+4y2−8y+4=1\Rightarrow 4{x^2} - 12x + 9 + 4{y^2} - 8y + 4 = 1

⇒4x2+4y2−12x−8y+12=0\Rightarrow 4{x^2} + 4{y^2} - 12x - 8y + 12 = 0

⇒x2+y2−3x−2y+3=0\Rightarrow {x^2} + {y^2} - 3x - 2y + 3 = 0

Radius = 94+1−3=12\sqrt {{9 \over 4} + 1 - 3} = {1 \over 2}

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