Question

The distance from the origins of centers of three circles ${x^2} + {y^2} - 2\lambda x = {c^2}$ (where $c$ is a constant and $\lambda$ is a variable ) are in geometrical progression; prove that the lengths of the tangents drawn to them from any point on the circle ${x^2} + {y^2} - {c^2} = 0$ are also in geometrical progression.

Answer 1

Hint:In this question let the tangents are drawn at any point ${\text{P}}\left( {h,k} \right)$ on ${x^2} + {y^2} - {c^2} = 0$ to all the three circles $\therefore {h^2} + {k^2} - {c^2} = 0$.

Complete step-by-step answer:
Given circles are,
${x^2} + {y^2} - 2{\lambda _1}x - {c^2} = 0$....take this as equation $(1)$
${x^2} + {y^2} - 2{\lambda _2}x - {c^2} = 0$....take this as equation $(2)$
${x^2} + {y^2} - 2{\lambda _3}x - {c^2} = 0$....take this as equation as $(3)$
Consider the distance from origin of centers are ${\lambda _1}$, ${\lambda _2}$ and ${\lambda _3}$
If a,b and c are in G.P then we can write $b^2=ac$
Similarly ${\lambda _1}$, ${\lambda _2}$ and ${\lambda _3}$ are in G.P as they given in question ,we can write
${\lambda _2}^2 = {\lambda _1}{\lambda _3}$
Let the tangents are drawn at any point ${\text{P}}\left( {h,k} \right)$ on ${x^2} + {y^2} - {c^2} = 0$ to all the three circles
$\therefore {h^2} + {k^2} - {c^2} = 0$....take this as equation $(4)$
Length of tangent from ${\text{P}}\left( {h,k} \right)$ to equation $(1)$ is:
${\text{P}}{{\text{T}}_1} = \sqrt {{h^2} + {k^2} - 2{\lambda _1}h - {c^2}} \\\ {\text{P}}{{\text{T}}_1} = \sqrt {{h^2} + {k^2} - {c^2} - 2{\lambda _1}h} \\\$
Substituting equation fourth we get:
${\text{P}}{{\text{T}}_1} = \sqrt {0 - 2{\lambda _1}h} \\\ {\text{P}}{{\text{T}}_1} = \sqrt { - 2{\lambda _1}h} \\\ {\text{P}}{{\text{T}}_1}^2 = - 2{\lambda _1}h \\\$
Similarly length of tangent from ${\text{P}}\left( {h,k} \right)$ to equation second is:
${\text{P}}{{\text{T}}_2}^2 = - 2{\lambda _2}h$
Similarly length of tangent from ${\text{P}}\left( {h,k} \right)$ to equation third is:
${\text{P}}{{\text{T}}_3}^2 = - 2{\lambda _3}h$
If lengths of tangents are in geometric progression then square of their length will also be in geometric progression
${\text{P}}{{\text{T}}_2}^4 = {\text{P}}{{\text{T}}_1}^2 \times {\text{P}}{{\text{T}}_3}^2$
substituting the values we get:
$\Rightarrow {\left( { - 2{\lambda _2}h} \right)^2} = - 2{\lambda _1}h \times - 2{\lambda _3}h \\\ \Rightarrow \lambda _2^2 = {\lambda _1}{\lambda _3} \\\$
Hence proved that their lengths are in geometric progression.

Note:- Here we considered the distance from origin of centers are ${\lambda _1}$, ${\lambda _2}$ and ${\lambda _3}$ and the length of tangent from point ${\text{P}}\left( {h,k} \right)$ to equation first is calculated to be $- 2{\lambda _1}h$ similarly for equation second and third is $- 2{\lambda _2}h$ and $- 2{\lambda _3}h$ since the lengths of tangents are in geometric progression then square of their length will also be in geometric progression therefore ${\text{P}}{{\text{T}}_2}^4 = {\text{P}}{{\text{T}}_1}^2 \times {\text{P}}{{\text{T}}_3}^2$ after substituting the values we got $\lambda _2^2 = {\lambda _1}{\lambda _3}$, hence proved that their lengths are in geometric progression.