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

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Solution

$The centre of the required circle is image of the the centre (3, 2) with respect to the line mirror x + y - 1 = 0 and radius equal to the radius of the given circle . Let (h, k) be the image of the point (3, 2) with respect to the mirror line . Then the mid point of the line joining C (3, 2) and P (h, k) lies on the line mirror . (\frac{h + 3}{2}) + (\frac{k + 2}{2}) - 1 = 0 h + k + 3 = 0 . . . . . . (1) Also CP is perpendicular to the x + y - 1 = 0 so product of slop will be - 1, \frac{k - 2}{h - 3} \times - 1 = - 1 or h - k = 1 . . . . . . (2) solving (1) and (2) we get, h = - 1 and k = - 2$

$thus the centre of the image circle is (- 1, - 2) . The radius of the image circle is same as the radius of x^{2} + y^{2} - 6 x - 4 y + 12 = 0 Radius = \sqrt{g^{2} + f^{2} - c} = \sqrt{3^{2} + 2^{2} - 12} = 1 unit Hence the equation of the required circle is, {(x + 1)}^{2} + {(y + 2)}^{2} = 1^{2} x^{2} + y^{2} + 2 x + 4 y + 4 = 0$

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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 isA) 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

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