$x^2 + y^2 + 2\lambda x + 5 =0 and x^2 + y^2 + 2\lambda y + 5 =0$ are two circles. P is a point on the line x–y=0. If PA and PB are the lengths of the tangents from P to the two cricles and PA = 3, then PB =

$$\begin{gathered} Given circles are \\ S\equiv x^2+y^2+2\lambda x+5=0\to \left(1\right) \\ and S\text{'}\equiv x^2+y^2+2\lambda y+5=0\to \left(2\right) \\ Radical axis of \left(1\right) and \left(2\right) is S-S\text{'}=0 \\ \Rightarrow x-y=0 \\ Since P is point on the line x-y=0 which is radical axis \\ and given PA=3 then PB=3 \end{gathered}$$