Inverse point of P(1,2) w.r.t. the circle is the foot of the perpendicular of P on the polar of P. Given circle is x2+y2−4x−6y+9=0 Polar of P(1,2) is x⋅1+y⋅2−2(x+1)−3(y+2)+9=0 ⇒x+2y−2x−2−3y−6+9=0 ⇒x+y−1=0 Verifying inverse point of P is (0,1).