Question

If $(1+\alpha x{)}^n=1+8x+24x^2+...$ and a line through $P(\alpha ,n)$ cuts the circle $x^2+y^2=4$ in $A$ and $B$, then $PA.PB =

• A. $4$
• B. $8$
• C. $16$
• D. $32$

Answer 1

The correct option is C $16$

Given $(1+ax{)}^n=1+8x+24x^2+...............$ ---(1)
we know, $(1+n{)}^n=n_{c_0}{x}^0+n_{c_1}{x}^1+n_{c_2}{x}^2+..........+n_{c_n}{x}^n$ ---(2)
so, by comparing (1) & (2)
$a{h}_{c_1}=8$ and $a{h}_{c_2}=24$
$h_{c_1}=\frac{\left(h\right)!}{(h-1)!1!}=h$
so, $a^h=8$ ---- (3) and $a^2(h-1)\left(h\right)=24\times 2=48$ ---- (4)
so, from (3) & (4),
$a^2(h^2-h)=48$
$(a^2{h}^2-ah\times a)=48$
$(64-8a)=48$
$a=2$ so $n=4$
so, $P(a,n)\equiv P(2,4)$
$P(2,4)$ lies outside circle $x^2+y^2=4$
line through $P(2,4)$ cuts $x^2+y^2=4$ at A and B
so, By property $PA\cdot PB=P{T}^2=(\sqrt{2^2+4^2-4}{)}^2$
$=16$