Question

Two circles with radii '$a$' and '$b$' respectively touch each other externally. Let '$c$' be the radius of a circle that touches these two circles as well as a common tangent to the two circles. Then

• A. $-\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}=\frac{1}{\sqrt{c}}$
• B. $\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}=-\frac{1}{\sqrt{c}}$
• C. $\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}=\frac{1}{\sqrt{c}}$
• D. $\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}=\frac{1}{\sqrt{c}}$

Answer 1

The correct option is C $\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}=\frac{1}{\sqrt{c}}$

$PR=MC=\sqrt{A{C}^2-A{M}^2}$
$\therefore \sqrt{(a+c{)}^2-(a-c{)}^2}=2\sqrt{ac}$
Similarly, $QR=2\sqrt{b}c$
Now, $PQ=PR=RQ=2\sqrt{ac}+2\sqrt{bc}.....\left(1\right)$
Draw PN parallel to AB
$\therefore PN=AB=a+b,$
$QN=BQ-BN=b-a$
$\therefore P{Q}^2=P{N}^2-Q{N}^2$
$=(a+b{)}^2-(a-b{)}^2=4ab$
$\Rightarrow PQ=2\sqrt{ab}.....\left(2\right)$
$\therefore$ From (i) and (ii)
$\Rightarrow \frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}+\frac{1}{b}.$