Question

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

• A. $x^2+y^2$ + 2x + 4y + 4 = 0
• B. $x^2+y^2$ - 2x + 4y + 4 = 0
• C. $x^2+y^2$ + 2x + 4y - 4 = 0
• D. $x^2+y^2$ + 2x - 4y - 4 = 0

Answer 1

The correct option is A $x^2+y^2$ + 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
$\therefore$ 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