Question

If the distances from the origin of the centres of three circles $x^2+y^2+2{\lambda }_{i}x-c^2=0(i=1,2,3)$ are in GP, then the length of the tangents drawn to them from any point on the circle $x^2+y^2=c^2$ are in?

• A. AP
• B. GP
• C. HP
• D. None of these

Answer 1

The correct option is B GP

$\text{Solve:-}{x}^2+y^2+2{\lambda }_{i}x-c^2=0(i=1,2,3)$

$\text{distance of centre of circle}\left(c_1\right)\text{from}$

$\text{origin}is={\lambda }_1$

Similaty, distance of circle $\left(c_2\right)={\lambda }_2$

and distance of centre of $\text{circle}\left(c_3\right)={\lambda }_3$

according to condition ${\lambda }_1,{\lambda }_2$ and ${\lambda }_3$ are in $GP$

$\Rightarrow {\lambda }_2{}^2={\lambda }_1{\lambda }_3-\left(1\right)$

We know that length of tangent from any Point is $=\sqrt{S_1}$

where $s\equiv x^2+y^2+2{\lambda }_{i}x-c^2$

length of tangent to $c_1=\sqrt{x^2+y^2+2{\lambda }_{1}x-c^2}$

$l_1=\sqrt{{\left(\frac{c}{\sqrt{2}}\right)}^2+{\left(\frac{c}{\sqrt{2}}\right)}^2+2{\lambda }_1\times \frac{c}{\sqrt{2}}-c^2}$

$\Rightarrow l_1=\sqrt{\sqrt{2}{\lambda }_{1}C}-\left(ii\right)$

similary, length of tangent on $c_2$ from

a paint $(\frac{c}{\sqrt{2}},\frac{c}{\sqrt{2}})$ on circle

$x^2+y^2=c^2$

is $=\sqrt{{\left(\frac{c}{\sqrt{2}}\right)}^2+{\left(\frac{c}{\sqrt{2}}\right)}^2+2{\lambda }_2\frac{c}{\sqrt{2}}-c^2}$

$\Rightarrow l_2=\sqrt{\sqrt{2}{\lambda }_{2}c}$ - (iii)

and, length of tangent on $c_3$ from a

point $(\frac{c}{\sqrt{2}},\frac{c}{\sqrt{2}})$ is

$\sqrt{{\left(\frac{c}{\sqrt{2}}\right)}^2+{\left(\frac{c}{\sqrt{2}}\right)}^2+2{\lambda }_3\frac{c}{\sqrt{2}}-c^2}$

$\Rightarrow l_3=\sqrt{\sqrt{2}{\lambda }_{3}c}-\left(iv\right)$

from (ii), (ii) and

(iv) we get

$l_2^2=l_1{l}_3$

so, length are in GP.