The equation of the image of the circle $x^{2} + y^{2} - 6 x - 4 y + 12 = 0$ by the line mirror $x + y - 1 = 0$ is

x^{2} + y^{2} + 2 x + 4 y + 4 = 0

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x^{2} + y^{2} - 2 x + 4 y + 4 = 0

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x^{2} + y^{2} + 2 x + 4 y - 4 = 0

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x^{2} + y^{2} + 2 x - 4 y + 4 = 0

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Solution

The correct option is A $x^{2} + y^{2} + 2 x + 4 y + 4 = 0$
The image of circle w.r.t same line means image of centre w.r.t that line without changing radius
$x^{2} + y^{2} - 6 x - 4 y + 12 = 0$
Centre $= (3, 2)$ Radius $= 1$
Image of $(3, 2)$ w.r.t $x + y - 1 = 0$
$\frac{x - 3}{1} = \frac{y - 2}{1} = - 2 \frac{(3 + 2 - 1)}{1^{1} + 1^{2}} = - 4$
$x = - 1$ , $y = - 2$
Then equation of image of circle is
$(x + 1)^{2} + (y + 2)^{2} = (1)^{2}$
$\Rightarrow x^{2} + y^{2} + 2 x + 4 y + 4 = 0$

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The equation of the image of the circle x2+y2−6x−4y+12=0 by the line mirror x+y−1=0 is

The equation of the image of the circle $x^{2} + y^{2} - 6 x - 4 y + 12 = 0$ by the line mirror $x + y - 1 = 0$ is