The equation of polar w.r.t the point (1,1) to the circle x2+y2−4x−6y+12=0 is x⋅1+y⋅1−2(x+1)−3(y+1)+12=0 ⇒−x−2y+7=0 ⇒x+2y−7=0 Since, the inverse of the point (1,1) is the foot (h,k) of the perpendicular from the point (1,1) to the line x+2y−7=0 So,