If the distances from the origin of the centers of three circles `x^2+y^2+2lambdax-c^2=0,(i=1,2,3),` are in GP, then prove that the lengths of the tangents drawn to them from any point on the circle `x^2+y^2=c^2` are in GP.

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

B

Explanation

The centres of the given circles are `(-lambda_(i), 0)` (i=1, 2, 3).The distances from the origin of the centres are `lambda-(1), lambda_(2)` and `lambda_(3)`.
Let P(h, k) be any point on the circle `x^(2)+y^(2)=c^(2)`. Then,
`h^(2)+k^(2)=c^(2)`
Now,
`L_(i)` = Length of the tangent from (h, k) to
`x^(2)+y^(2)+1lambda_(i)x x-c^(2)=0`
`rArr L_(i)=sqrt(h^(2)+k^(2)+2lambda_(i)h-c^(2))`
`rArr L_(i)=sqrt(c^(2)+2lambda_(i)h-c^(2))" " [:' h^(2)+k^(2)=c^(2)]`
`rArr L_(i)=sqrt(2 ,lambda_(i) h), " " i=1, 2, 3`.
`:. L_(2)^(2)=2lambda_(2)h`
`rArr L_(2)^(2)=2h(sqrt(2lambda_(1)lambda_(3))" " [:' lambda_(2)^(2)=lambda_(1)lambda_(3)]`
`rArr L_(2)^(2)=sqrt(2 lambda_(1)h)sqrt(2 lambda_(3)h)=L_(1)L_(3)`
Hence, `L_(1), L_(2), L_(3)` are in G.P.