Question

Three circles touch one another externally. If the tangents at their points of contact meet at a point whose distance from a point of contact is $4$, then the ratio of product of radii to the sum of the radii of the circles, is

• A. $2:1$
• B. $4:1$
• C. $8:1$
• D. $16:1$

Answer 1

The correct option is D $16:1$

Let $r_1,r_2,r_3$ be the radii of circles with centres $A,B,C,$ respectively. These circles touch one another externally.
Let $O$ be their radical centre.
Also, $O$ is the incentre if $△ABC$.
Since $OL=4$,
$\therefore$ the radius of the incircle is 4.
Now $AB=r_1+r_2=c$
$BC=r_2+r_3=a$
$AB=r_1+r_3=b$
$\therefore s=\frac{a+b+c}{2}=r_1+r_2+r_3$
So, $s-a=r_1,s-b=r_2$ and $s-c=r_3$
Thus, area of $△ABC=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(r_1+r_2+r_3)r_1{r}_2{r}_3}$
We know that,
$\text{radius of incircle}=\frac{\text{area of traingle ABC}}{\text{semiperimeter of triangle ABC}}$
$\therefore \frac{\sqrt{(r_1+r_2+r_3)r_1{r}_2{r}_3}}{r_1+r_2+r_3}=4$
$\Rightarrow \frac{r_1{r}_2{r}_3}{r_1+r_2+r_3}=16$
Therefore, the required radio is $16:1$.