Question

The equation of two circles are $x^2+y^2+2\lambda x+5=0$ and $x^2+y^2+2\lambda y+5=0$. $P$ is any point on the line $x-y=0$. If $PA$ and $PB$ are the lengths of the tangents from $P$ to the two circles and $PA=3$ then $PB$ is equal to

• A. $1.5$
• B. $6$
• C. $3$
• D. none of these

Answer 1

Answer: (C)

$3$

$S_1\equiv x^2+y^2+2\lambda x+5=0$
$S_2\equiv x^2+y^2+2\lambda y+5=0$
equation of radical axis = $S_1-S_2\Rightarrow y-x=0$
Clearly line $y-x=0$ is radical axis of the given circles.
Hence length of tangents will be same. $PA=PB=3$
$x^2+y^2+2\lambda x+5=0$ and $x^2+y^2+2\lambda y+5=0$