Question

A line passing through $P(1,2)$ inclined at an angle $\theta$ with positive $x-$axis cuts the circle $x^2+y^2-6x-8y+24=0$ at $A$ and $B$ such that $PA+PB=4\sqrt{2}.$ Then the value of $\cos \theta +\sin \theta$ is

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

Answer 1

The correct option is A $\sqrt{2}$


For $P(1,2),$
$S_1=1+4-6-16+24=7>0$
$\therefore P$ lies outside the circle $x^2+y^2-6x-8y+24=0$
Equation of line $L$ is $y-2=\tan \theta (x-1)$
Let $A$ be the point on $L$ at a distance of $r$ from $P.$
Then $A\equiv (1+r\cos \theta ,2+r\sin \theta )$
$\therefore {(1+r\cos \theta )}^2+{(2+r\sin \theta )}^2-6(1+r\cos \theta )-8(2+r\sin \theta )+24=0$
$\Rightarrow r^2-4r(\cos \theta +\sin \theta )+6=0$
Let $r_1,r_2$ be roots of the equation.
Then $r_1+r_2=4(\cos \theta +\sin \theta )$
Given, $PA+PB=r_1+r_2=4\sqrt{2}$
$\therefore \cos \theta +\sin \theta =\sqrt{2}$