Question

Two circle with radii r₁ and r₂ touch each other externally. Let r be the radius of a circle which touches these two circle as well as a common tangent to the two circles, Prove that: ${\displaystyle \frac{1}{\sqrt{r}}+\frac{1}{{\sqrt{r}}_1}+\frac{1}{{\sqrt{r}}_2}}$.

Answer 1

From the adjoining figure,
PQ = SY = ${\displaystyle \sqrt{{\text{XY}}^2-{\text{XS}}^2}}$
= ${\displaystyle \sqrt{{(r_1+r_2)}^2-{(r_1-r_2)}^2}}$
= ${\displaystyle \sqrt{4r_1{r}_2}}$
= ${\displaystyle \sqrt{r_1{r}_2}}$

Similarly, PR = ${\displaystyle 2\sqrt{r{r}_1}}$ and RQ = ${\displaystyle 2\sqrt{r{r}_2}}$
Now, PQ = PR + RQ
${\displaystyle 2\sqrt{r_1{r}_2}=2\sqrt{r{r}_1}=2\sqrt{r{r}_2}}$

⇒ ${\displaystyle \sqrt{r_1{r}_2}=\sqrt{r{r}_1}=\sqrt{r{r}_2}}$

Dividing by ${\displaystyle \sqrt{r{r}_1{r}_2}}$ on both sides,

⇒ ${\displaystyle \frac{1}{\sqrt{r}}+\frac{1}{{\sqrt{r}}_1}+\frac{1}{{\sqrt{r}}_2}}$.
Hence proved.