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Circle question

2021 · 26 Feb · Shift 2 · Q37
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  5. /2021 · 26 Feb · Shift 2 · Q37

Circle question

2021 · 26 Feb · Shift 2 · Q37

JEE MainMathematicsCircleMCQ+4 / −1
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 :
  1. A
    14{1 \over 4}41​
  2. B
    12{1 \over 2}21​
  3. C
    1
  4. D
    13{1 \over 3}31​
View written solutionFree

Correct answer: B

  1. Let a variable point on the circle x2+y2=1x^2+y^2=1x2+y2=1 be P(a,b)P(a,b)P(a,b).

    Since PPP lies on the circle, a2+b2=1.a^2+b^2=1.a2+b2=1.

  2. The fixed point is A(3,2)A(3,2)A(3,2).

    Let the midpoint of APAPAP be M(x,y)M(x,y)M(x,y).

    Using midpoint formula, x=3+a2,y=2+b2.x=\frac{3+a}{2},\qquad y=\frac{2+b}{2}.x=23+a​,y=22+b​.

  3. Express a,ba,ba,b in terms of x,yx,yx,y: a=2x−3,b=2y−2.a=2x-3,\qquad b=2y-2.a=2x−3,b=2y−2.

  4. Substitute into the circle condition a2+b2=1a^2+b^2=1a2+b2=1: (2x−3)2+(2y−2)2=1.(2x-3)^2+(2y-2)^2=1.(2x−3)2+(2y−2)2=1.

    Expanding, 4x2−12x+9+4y2−8y+4=1,4x^2-12x+9+4y^2-8y+4=1,4x2−12x+9+4y2−8y+4=1, 4x2+4y2−12x−8y+12=0,4x^2+4y^2-12x-8y+12=0,4x2+4y2−12x−8y+12=0, x2+y2−3x−2y+3=0.x^2+y^2-3x-2y+3=0.x2+y2−3x−2y+3=0.

  5. Complete the squares: x2−3x+y2−2y+3=0,x^2-3x+y^2-2y+3=0,x2−3x+y2−2y+3=0, (x−32)2−94+(y−1)2−1+3=0,\left(x-\frac{3}{2}\right)^2-\frac{9}{4}+\left(y-1\right)^2-1+3=0,(x−23​)2−49​+(y−1)2−1+3=0, (x−32)2+(y−1)2=14.\left(x-\frac{3}{2}\right)^2+\left(y-1\right)^2=\frac{1}{4}.(x−23​)2+(y−1)2=41​.

  6. Hence the locus is a circle with radius r=14=12.r=\sqrt{\frac{1}{4}}=\frac{1}{2}.r=41​​=21​.

  7. Comparing with the options, the correct option is B.

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