Two circles with the radii $a$  and $b$  units touches externally. A circle with the radius ‘c’ touches both circles and the common tangent to the circles too. Then prove that $\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}=$

We know that , Radius and Tangent both are perpendicular at point of Contact

$\therefore OX\perp XY;MZ\perp XY,OY\perp XY$

$\therefore$ Draw PQ||ZX

Such that, $PX=MZ=QY=C$

$\therefore OP=a-c;O\text{'}Q=b-c$

$OM=a+c;O\text{'}M=b+c$

$PQ=PM+MQ$

$=\sqrt{{\left(a+c\right)}^2-{\left(a-c\right)}^2}+\sqrt{{\left(b+c\right)}^2-{\left(b-c\right)}^2}$

$=\sqrt{4ac}+\sqrt{4bc}$

But $PQ=XY$

$\therefore XY=\sqrt{4ac}+\sqrt{4bc}$

But , $\sqrt{{\left(a+b\right)}^2-{\left(b-a\right)}^2}=XY=\sqrt{4ab}$

$\therefore \sqrt{4ac}+\sqrt{4bc}=\sqrt{4ab}$

$\sqrt{ac}+\sqrt{bc}=\sqrt{ab}$

$\sqrt{\frac{c}{b}}+\sqrt{\frac{c}{a}}=1$

$\Rightarrow \frac{1}{\sqrt{c}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}$