Question

When two circle neither intersect nor touch and are far apart, then four common tangents can be drawn to two circles, two of which are direct other two are transverse. Let the distance between the centres of the circles be ' $d$ ' and their radii are $r_1$ and $r_2$. The lengths of the direct common tangents and the transverse common tangents are given by $L_D=\sqrt{d^2-{\left(r_1-r_2\right)}^2}$ and $L_T=\sqrt{d^2{\left(r_1+r_2\right)}^2}$ respectively.
Two circles of radii $a$ and $b(a>b)$ touch each other externally. The radius of a circle, which touches both the circles externally and also their common tangents, is equal to :

• A. $\frac{a+b}{2}$
• B. $\frac{a^2+b^2}{a-b}$
• C. $\frac{ab}{(\sqrt{a}+\sqrt{b}{)}^2}$
• D. $\frac{ab}{a+b}$

Answer 1

Answer: (C)

Let the desired radius be ' $r$ ', then
$AB=\sqrt{(r+a{)}^2-(r-a{)}^2}=2\sqrt{ra}$
$BC=\sqrt{(r+b{)}^2-(r-b{)}^2}=2\sqrt{rb}$
$AC=\sqrt{(a+b{)}^2-(a-b{)}^2}=2\sqrt{ab}$

Now, $AB+BC=AC$
$\Rightarrow r=\frac{ab}{(\sqrt{a}+\sqrt{b}{)}^2}$