The equation of the circle, which is the mirror image of the circle, in the line is ml o`Ùk dk lehdj.k] tks o`Ùk esa niZ.k izfrfcac gS]

JEE MainMathematicsModerateCircles

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

Show answer and solution

Centre of circle (1, 0), radius = 1
Image of (1, 0) w.r.t. Line x + y - 3 = 0 is

\frac{x - 1}{1} = \frac{y - 0}{1} = \frac{- 2(1 + 0 - 3)}{1 + 1} = 2

\Rightarrow x = 3,y = 2

Now circle with centre (3, 2) and radius 1 is

(x - 3)^{2} + (y - 2)^{2} = 1

\Rightarrow x^{2} + 9 - 6x + y^{2} + 4 - 4y = 1 \Rightarrow x^{2} + y^{2} - 6x - 4y + 12 = 0

o`Ùk Unz (1, 0), f=T;k = 1

\frac{x - 1}{1} = \frac{y - 0}{1} = \frac{- 2(1 + 0 - 3)}{1 + 1} = 2 \Rightarrow x = 3,y = 2

dsUnz(3, 2)rFkk f=T;k1 okyk o`Ùk

(x - 3)^{2} + (y - 2)^{2} = 1

\Rightarrow x^{2} + 9 - 6x + y^{2} + 4 - 4y = 1 \Rightarrow x^{2} + y^{2} - 6x - 4y + 12 = 0

Source: Grass-G / M-27 Mathematics Paper 2 15 Jan.docx

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