IIT JEE Advanced 2012 Paper 1

Equation of the chord bisected at P(h, k)
hx + ky = h2+k2 (1)
Let any point on line be (α,

4

5

(α−4)).
Equation of the chord of contact is
αx+(

4

5

(α−4))y = 9 (2)
Comparing Eqs. (1) and (2), we get

h

α

=

k

4 5 (α−4)

=

h2+k2

9

α =

20h

4h−5k

Now,

h(4h−5k)

20h

=

h2+k2

9

20(h2+k2) = 9 (5h - 5k)
20(x2+y2)−36x+45y=0