The equation of pole w.r.t. the point (1,2) to the circle x2+y2−4x−6y+9=0 is x+2y−2(x+1)−3(y+2)+9‌=0 ⇒ x+y−1‌=0 Since, the inverse of the point (1,2) is the foot (α,β) of the perpendicular from the point (1,2) to the line x+y−1. ‌∴‌‌
α−1
1
=‌
β−2
1
‌=−‌
(1⋅1+1⋅2−1)
12+12
⇒‌α−1‌=β−2=−1 ⇒‌α‌=0,β=1 Hence, required point is (0,1).
