If the chord of contact of the tangents drawn from a point on the circle `x^2+y^2+y^2=a^2` to the circle `x^2+y^2=b^2` touches the circle `x^2+y^2=c^2` , then prove that `a ,b` and `c` are in GP.

Options

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

Answer

B

Explanation

Let `P(x_(1), y_(1))` be a point on the circle `x^(2)+y^(2)=a^(2)`
Then,
`x_(1)^(2)+y_(1)^(2)=a^(2) " " ...(i)`
The equation of the chord of contact of tangents drawn from `(x_(1), y_(1))` to the circle `x^(2)+y^(2)=b^(2)` is
`x x_(1)+y y_(1)=b^(2) " " ...(ii)`
This touches the circle `x^(2)+y^(2)=c^(2)`
`:.` Length of the perpendicular from (0, 0) to line (ii) is equal to the radius of `x^(2)+y^(2)=c^(2)`

`rArr|(0xx x_(1)+0xxy_(1)-b^(2))/(sqrt(x_(1)^(2)+y_(1)^(2)))|=c`
`rArr(b^(2))/(sqrt(a^(2)))=c" " `[Using (i)]
`rArrb^(2)=ac rArr a, b, c` are in G.P.