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NDA 2025 Maths · Mathematics · Circles · medium

The equation of the circle which is concentric with x^2+y^2-4x+6y-3=0 and passes through the point (1,1) is

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The equation of the circle which is concentric with x2+y2−4x+6y−3=0x^2+y^2-4x+6y-3=0 and passes through the point (1,1)(1,1) is
  1. A. (x−2)2+(y+3)2=16(x-2)^2+(y+3)^2=16
  2. B. (x−2)2+(y+3)2=17(x-2)^2+(y+3)^2=17
  3. C. (x+2)2+(y−3)2=17(x+2)^2+(y-3)^2=17
  4. D. (x−2)2+(y+3)2=25(x-2)^2+(y+3)^2=25
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Answer: (B) (x−2)2+(y+3)2=17(x-2)^2+(y+3)^2=17

The given circle has centre (2,−3)(2,-3). A concentric circle has the same centre, and its radius is the distance to (1,1)(1,1): r2=(1−2)2+(1+3)2=17r^2=(1-2)^2+(1+3)^2=17. The option with 1616 keeps the radius of the original circle (r2=4+9+3=16r^2=4+9+3=16), which would not pass through (1,1)(1,1), and (x+2)2+(y−3)2(x+2)^2+(y-3)^2 has the centre's signs reversed.
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Original question written in the NDA 2025 Maths pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake