Three circles touch each other externally. The tangents at their point of contact meet at a point whose distance from a point of contact is 4. Then, the ratio of their product of radii to the sum of the radii is

Answer

`16 : 1`

Explanation

Suppose the circle have centres at `C_(1), C_(2) and C_(3)` with Radius `R_(1), R_(2) and R_(3)` respectively.Let the circle touch at A,B and C. Let the common tangents at A,B and C meet at O. We have OA = Ob = OC = 4 [given] Now, the circle with centre at O and passing through A, B and C is the incircle of the triangle `C_(1)C_(2)C_(3)` (because`OAbot C_(1)C_(2)`).
Therefore, the inradius of `Delta C_(1)C_(2)C_(3)` is 4 .
and `r=(Delta)/(s)" "...(i)`

Nowm perimeter of a triangle
`2s=R_(1)+R_(2)+R_(2)+R_(3)+R_(3)+R_(1)`
`rArr 2s=2(R_(1)+R_(2)+R_(3))`
`rArr s=R_(1) + R_(2)+R_(3)`
and `Delta = sqrt(s(s-a)(s-b)(s-c))`
`=sqrt((R_(1)+R_(2)+R_(3))(R_(3))(R_(2))(R_(1)))`
From Eq. (i), `4 =sqrt((R_(1)R_(2)R_(3))(R_(1)+R_(2)+R_(3)))/(R_(1)+R_(2)+R_(3))`
`rArr 16 =sqrt(R_(1)R_(2)R_(3)(R_(1)+R_(2)+R_(3)))/(R_(1)+R_(2)+R_(3))`
`rArr 16 = (R_(1)R_(2)R_(3))/(R_(1)+R_(2)+R_(3))`