Question

A circle $C_1$ of radius $b$ touches the circle $x^2+y^2=a^2$ externally and has its centre on the positive xaxis; another circle $C_2$ of radius $c$ touches the circle $C_1$ externally and has its centre on the positive x-axis. Given $a<b<c$, then the three circles have a common tangent if $a,b,c$ are in

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

Answer 1

Answer: (B)

$G.P.$

The centre of $C_1$ is $(a+b,0)$ and the centre of $C_2$ is $(a+2b+c,0)$
Let $y=mx+k$ be a tangent common to the three circles.Since it touches $x^2+y^2=a^2,$ $C_1$ and $C_1$
$\frac{k}{\sqrt{1+m^2}}=\pm a,\frac{m(a+b)+k}{\sqrt{1+m^2}}=\pm b$ and
$\frac{m(a+2b+c)+k}{\sqrt{1+m^2}}=\pm c$
As the centre of the three circles lie on the same side of the line $y=mx+k,$ taking the same sign, say positive, in the three relations we get,
$\frac{k}{\sqrt{1+m^2}}=a=b-m(a+b)=c-m(a+2b+c)$
$\frac{a+b}{b-a}=\frac{a+2b+c}{c-a}$ (eliminating m)
$\Rightarrow (a+b)(c-a)=(b-a)(a+2b+c)$
$\Rightarrow ac-a^2+bc-ba=ba-a^2+2b^2-2ab+bc-ac$
$\Rightarrow 2ac=2b^2\Rightarrow ac=b^2$
$\Rightarrow a,b,c$ are in $G.P.$

Ans: B