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JEE MainMathematicsToughCircles

The equation of the circle, which is the mirror image of the
circle, x2+y2−2x=0x^{2} + y^{2} - 2x = 0, in the line, y = 3-x is
ml o`Ùk dk lehdj.k] tks o`Ùk x2 + y2 - 2x = 0

Show answer and solution
Centre of circle o`Ùk Unz (1, 0), radius f=T;k =1
Image of (1, 0) w.r.t. Line x + y-3 = 0 is
x−11=y−01=−2(1+0−3)1+1=2\frac{x - 1}{1} = \frac{y - 0}{1} = \frac{- 2(1 + 0 - 3)}{1 + 1} = 2
⇒x=3,y=2\Rightarrow x = 3,y = 2
Now circle with centre (3,2) and radius 1 is
(x−3)2+(y−2)2=1(x - 3)^{2} + (y - 2)^{2} = 1
dsUnz (3, 2) rFkk f=T;k 1 okyk o`Ùk (x−3)2+(y−2)2=1(x - 3)^{2} + (y - 2)^{2} = 1
⇒x2+9−6x+y2+4−4y=1\Rightarrow x^{2} + 9 - 6x + y^{2} + 4 - 4y = 1
⇒x2+y2−6x−4y+12=0\Rightarrow x^{2} + y^{2} - 6x - 4y + 12 = 0

Source: Grass-G / M-25 Mathematics Paper 7 Jan.docx

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