Question

Let a line be drawn through the fixed point $P(a,b)$ to cut the circle $x^2+y^2=r^2$ at $A$ and $B.$ If $PT$ is the length of tangent drawn from $P,$ then

• A. $PA,PT,PB$ are in $G.P.$
• B. $PA.PB=a^2+b^2–r^2$
• C. $PA.PB=$ constant
• D. $PA.PB=r^2$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $PA,PT,PB$ are in $G.P.$
B $PA.PB=a^2+b^2–r^2$
C $PA.PB=$ constant

Equation of line through $(a,b)$ is

$\frac{x-a}{\cos \theta }=\frac{y-b}{\sin \theta }=\lambda$ ... (1)

$\lambda$ is the distance of $(x,y)$ from $P(a,b)$

$(\lambda \cos \theta +a,\lambda \sin \theta +b)$ lies on the circle $x^2+y^2=r^2$

${\lambda }^2+2\lambda (a\cos \theta +b\sin \theta )+a^2+b^2-r^2=0$ ... (2)

line (1) meets the circle in $A$ and $B$ then $PA$ and $PB$ are the roots of equation (2). So,

$PA.PB=a^2+b^2-r^2=(PT{)}^2$

$[\because (PT{)}^2=a^2+b^2-r^2]$

Hence, $PA,PT,PB$ are in G.P. and

$PA.PB=$ constant