Question

If the distances from the origin of the centre of the three circles $C_i:x^2+y^2-2a_{i}x=b^2(i=1,2,3a_i\in N)$ are in G.P. Let the length of the tengents drawn to $C_1,C_2\&C_3$ from any point on the curve $x+\sqrt{b^2-y^2}=0$ are $l_1,l_2\&l_3$ respectively, then

• A. $2l_2=l_1+l_3$
• B. $l_2=\sqrt{l_1{l}_3}$
• C. $l_2=\frac{2l_1{l}_3}{l_1+l_3}$
• D. $l_2^2=l_1^2+l_3^2$

Answer 1

The correct option is B $l_2=\sqrt{l_1{l}_3}$

Any point on the circle $x^2+y^2=b^2$ is $(b\cos \theta ,b\sin \theta )$, where $\theta \in (\frac{\pi }{2},\frac{3\pi }{2})$
Given condition $a_2^2=a_1{a}_3$
Let the lengths of the tangents drawn from the circle $x^2+y^2=b^2$ to circles having centers $(a_1,0),(a_2,0),(a_3,0)$ are $l_1,l_2,l_3$ respectively

$\therefore l_1=\sqrt{|-2a_{1}b\cos \theta |}{l}_2=\sqrt{|-2a_{2}b\cos \theta |}{l}_3=\sqrt{|-2a_{3}b\cos \theta |}{l}_1{l}_3=\sqrt{4a_1{a}_3{b}^2{\cos }^2\theta }=|2a_{2}b\cos \theta |=l_2^2$