Circles Part 2

The given line is polar or P(2,β) w.r.t. given circle
x2+y2−4x−6y−3=0
Chord or contact
‌αx+βy−2(x+α)−3(y+β)−3=0
‌⇒(α−2)x+(β−3)y−(2α+3β+3)=0 . . . (i)
∵ But the equation of chord of contact is given
as : x+y−3=0 . . . (ii)
comparing the coefficients
‌

α−2

1

=‌

β−3

1

=−(‌

2α+3β+3

−3

)
On solving α=−6
β=−5
Now ‌‌4α−7β=11