Question

From the point $P(2,1),$ a line of slope $m\in R$ is drawn so as to cut the circle $x^2+y^2=1$ in points $A$ and $B$. If the slope $m$ is varied, then the greatest possible value of $PA+PB$ is

• A. $\frac{2}{\sqrt{5}}$
• B. $\frac{10}{\sqrt{5}}$
• C. $2\sqrt{5}$
• D. $\frac{1}{\sqrt{5}}$

Answer 1

The correct option is C $2\sqrt{5}$

Using parametric form,
let $(2+r\cos \theta ,1+r\sin \theta )$ be any point on line through $P$ having slope $m=\tan \theta$.
This point lies on the circle $x^2+y^2=1$
$\Rightarrow (2+r\cos \theta {)}^2+(1+r\sin \theta {)}^2=1$
We get quadratic equation $r^2+r(4\cos \theta +2\sin \theta )+4=0$
$\therefore PA+PB=|-4\cos \theta -2\sin \theta |\le 2\sqrt{5}$