EAMCET Engineering 2008 Solved Paper

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).