A line drawn through the point $P(4, 7)$cuts the circle  $x^2+y^2=9$ at the points A and B. Then $PA.PB$ is equal to 

we have 

$x^2+y^2=9$

Let line through P, A and B make angle $\theta$ with x-axis

$\therefore$ The equation of line 

$\frac{x-4}{\cos \theta }=\frac{y-7}{\sin \theta }=k\left(\text{ say }\right)$

Any point on this line is  $(k\cos \theta +4,k\sin \theta +7)$

This point will also lie on the circle.  

$\therefore (k\cos \theta +4{)}^2+(k\sin \theta +7{)}^2=9$

$\Rightarrow k^2{\cos }^2\theta +16+8k\cos \theta +k^2{\sin }^2\theta +49+14k\sin \theta =9$

$\Rightarrow k^2+k(8\cos \theta +14\sin \theta )+56=0$

$\therefore k_1\times k_2=PA\cdot PB=\frac{56}{1}=56$