Question:

Equation of the circle, which is the mirror image of the circle \(x^{2}+y^{2}-2x=0\) with respect to the line \(x+y=2\), is :

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The mirror image circle has the same radius; reflect the centre in the line.
Updated On: Nov 12, 2026
  • \(x^{2}+y^{2}-2x+4y+3=0\)
  • \(x^{2}+y^{2}-4x-2y+4=0\)
  • \(2(x^{2}+y^{2})-x+y+1=0\)
  • \(x^{2}+y^{2}-2x-4y+3=0\)
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The Correct Option is B

Solution and Explanation

The circle \(x^2+y^2-2x=0\) has centre (1, 0) and radius 1.
Reflection of a point (p, q) in the line \(x+y=2\) is \((2-q,\,2-p)\). So the image of (1, 0) is (2, 1).
The mirror circle has centre (2, 1) and radius 1: \((x-2)^2+(y-1)^2=1\).
Expanding: \(x^2+y^2-4x-2y+4=0\).
Answer: (2) \(x^{2}+y^{2}-4x-2y+4=0\)
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