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.