Question
A line is drawn through a fixed point $$P\left( {\alpha ,\,\beta } \right)$$ to cut the circle $${x^2} + {y^2} = {a^2}$$ at $$A$$ and $$B,$$ then $$PA.PB$$ is equal to :
A.
$${\alpha ^2} + {\beta ^2}$$
B.
$${\alpha ^2} + {\beta ^2} - {a^2}$$
C.
$${a^2}$$
D.
$${\alpha ^2} + {\beta ^2} + {a^2}$$
Answer :
$${\alpha ^2} + {\beta ^2} - {a^2}$$
Solution :
Any point on the line at a distance $$r$$ from the point $$P\left( {\alpha ,\,\beta } \right)$$ is $$\left( {\alpha + r\,\cos \,\theta ,\,\beta + r\,\sin \,\theta } \right)$$
If this point lies on $${x^2} + {y^2} = {a^2},$$ then
$$\eqalign{
& {\alpha ^2} + {r^2}{\cos ^2}\theta + 2\alpha r\,\cos \,\theta + {\beta ^2} + {r^2}{\sin ^2}\theta + 2\beta r\,\sin \,\theta = {a^2} \cr
& \Rightarrow {r^2} + 2r\left( {\alpha \,\cos \,\theta + \beta \,\sin \,\theta } \right) + {\alpha ^2} + {\beta ^2} = {a^2} \cr
& \Rightarrow {r^2} + 2r\left( {\alpha \,\cos \,\theta + \beta \,\sin \,\theta } \right) + {\alpha ^2} + {\beta ^2} - {a^2} = 0 \cr} $$
Now, if $$PA = {r_1}$$ and $$PB = {r_2},$$ then $${r_1}$$ and $${r_2}$$ must be roots of this equation.
$$\therefore \,PA.PB = {r_1}.{r_2} = {\alpha ^2} + {\beta ^2} - {a^2}$$