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

+ 2x + 4y + 4 = 0

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- 2x + 4y + 4 = 0

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+ 2x + 4y - 4 = 0

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+ 2x - 4y - 4 = 0

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Solution

The correct option is A + 2x + 4y + 4 = 0
For the circle $x^{2} + y^{2}$ - 6x - 4y + 12 = 0.
This circle has centre C(3, 2) and radius (r) = 1
Image of (3, 2) w.r.t the line x + y - 1 = 0 = (-1, -2)
Equation of the circle with centre (-1, -2) and radius 1 is,
$(x + 1)^{2} + (y + 2)^{2}$ = 1
$∴$ Image of $x^{2} + y^{2}$ - 6x - 4y + 12 = 0 w.r.t x + y - 1 = 0 is,
$x^{2} + y^{2}$ + 2x + 4y + 4 = 0

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

A) x^2+y^2+2x+4y+4=0 B) x^2+y^2+x+y=0 C) x^2+y^2-2x-4y+4=0 D) x^2+y^2-2x-4y-4=0

C_{1} : x^{2} + y^{2} - 2 x - 4 y - 4 = 0

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

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