Question

A line is drawn through a fixed point $P(\alpha ,\beta )$ to cut the $x^2+y^2=r^2$ at $A$ and $B$. Then, $PA$. $PB$ is equal to

• A. ${(\alpha +\beta )}^2-r^2$
• B. ${\alpha }^2+{\beta }^2-r^2$
• C. ${(\alpha -\beta )}^2+r^2$
• D. ${\alpha }^2-{\beta }^2$

Answer 1

The correct option is B ${\alpha }^2+{\beta }^2-r^2$

Co-ordinate of $A$ is

$\frac{x-\alpha }{\cos \theta }=\frac{y-\beta }{\sin \theta }=p$

Now,

$\Rightarrow$ $x-\alpha =p\cos \theta$

$\Rightarrow$ $x=\alpha +p\cos \theta$

Also,

$\Rightarrow$ $y-\beta =p\sin \theta$

$\Rightarrow$ $y=\beta +p\sin \theta$

So, we get $(x,y)=(\alpha +p\cos \theta ,\beta +p\sin \theta )$

$(\alpha +p\cos \theta ,\beta +p\sin \theta )$ lie on $x^2+y^2=r^2$

$\Rightarrow$ $(\alpha +p\cos \theta {)}^2$ $+(\beta +p\sin \theta {)}^2=r^2$

$\Rightarrow$ ${\alpha }^2+2\alpha pcos\theta +p^2{\cos }^2\theta +{\beta }^2+2\beta p\sin \theta +p^2{\sin }^2\theta =r^2$

$\Rightarrow$ $p^2+2p(\alpha \cos \theta +\beta \sin \theta )+{\alpha }^2+{\beta }^2-r^2=0$

$\Rightarrow$ $p_1{p}_0=PA.PB={\alpha }^2+{\beta }^2-r^2$