# If the distances from origin of centre of three circles $x^2+y^2+2\lambda_i-c^2=0$ (i=1,2,3) are in G.P…

Problem : If the distances from origin to the centre of three circles $x^2+y^2+2\lambda_ix-c^2=0$ (i=1,2,3) are in G.P ( Geometric progression). Then lengths of tangents drawn to them from any point on the circle $x^2+y^2=c^2$ are in

(a) G.P

(b) A.P

(c) H.P.

Solution : Here centre of the first circle is $(-\lambda_1,0)$...(i) ;

centre of the second circle is $(-\lambda_2,0)$...(ii) ; centre of the third circle is $(-\lambda_3,0)..(iii)$

Distance from the origin from these centres (i), (ii) \& (iii) are :

$(\lambda_1, \lambda_2,\lambda_3)$ respectively. As per the given condition : $\lambda_2^2 = \lambda_1\lambda_3$ ( as they are in G.P )

Let the point on the circle $x^2+y^2=c^2$ is $(\alpha, \beta)$ Now the distance from this point ie. $(\alpha,\beta)$ from the given three circles are given by :

$\alpha^2+\beta^2+2\lambda_1\alpha -c^2$;

$\alpha^2+\beta^2+2\lambda_2\alpha -c^2$

$\alpha^2+\beta^2+2\lambda_3\alpha -c^2$

Please suggest how to proceed on this further I am not getting any clue.. Thanks.

As $(\alpha,\beta)$ lies on the circle $x^2+y^2=c^2,$
$$\alpha^2+\beta^2=c^2\implies \alpha^2+\beta^2+2\lambda_1\alpha-c^2=?$$
• @user108258, So $$\alpha^2+\beta^2+2\lambda_i\alpha-c^2=2\lambda_1\alpha_i,$$ Now if $$\lambda_1,\lambda_2,\lambda_3$$ are in G.P.,so will be $2\lambda_1\alpha_i$-s – lab bhattacharjee Jul 18 '14 at 3:15