If the squares of the lengths of the tangents from a point p to the circles $x^2+y^2=a^2,x^2+y^2=b^2,x^2+y^2=c^2$ are in A.P then $a^2,b^2,c^2$ are in 

$S_{11}=x_1^2+y_1^2-a^2,S_{11}^1=x_1^2+y_1^2-b^2,S_{11}^{11}=x_1^2+y_1^2-c^2$

Since $S_{11},S_{11}^1,S_{11}^{11}$ are in A.P $\Rightarrow 2S_{11}^1=S_{11}+S_{11}^{11}$

$\Rightarrow 2x_1^2+2y_1^2-2b^2=x_1^2+y_1^2-a^2+x_1^2+y_1^2-c^2$

$-2b^2=-a^2-c^2$

$\Rightarrow a^2+c^2=2b^2$

$\therefore a^2,b^2,c^2$  are in A.P