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 Share with your friends Share 15 Anuradha Sharma answered this 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.h+32+k+22-1=0h+k+3=0......(1)Also CP is perpendicular to the x+y-1=0 so product of slop will be -1,k-2h-3×-1=-1or 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 x2+y2-6x-4y+12=0Radius = g2+f2-c=32+22-12=1 unitHence the equation of the required circle is, x+12+y+22=12x2+y2+2x+4y+4=0 19 View Full Answer
