In the figure (ii) given below, equal circles with centres O and O' touch each other at X. OO' is produced to meet a circle O' at A. AC is tangent to the circle whose centre is O. O'D is perpendicular to AC. Find the value of

(i) $\dfrac{\text{AO}'}{\text{AO}}$

(ii) $\dfrac{\text{area of △ADO}'}{\text{area of △ACO}}.$

From figure,

OC is radius and AC is tangent, then OC ⊥ AC.

Let radius of each equal circle = r.

(i) From figure,

AO = AO' + O'X + XO and AO' = O'X = XO = r (radius of circle)

AO = r + r + r = 3r.

$\therefore \dfrac{\text{AO}'}{\text{AO}} = \dfrac{r}{3r} = \dfrac{1}{3}.$

Hence, the value of $\dfrac{AO'}{AO} = \dfrac{1}{3}$.

(ii) Considering △ADO' and △ACO

∠A = ∠A (Common angles)

∠D = ∠C (Both are equal to 90°)

∴ By AA axiom △ADO' ~ △ACO.

Since triangles are similar hence the ratio of their areas is equal to the ratio of the square of the corresponding sides.

$\dfrac{\text{Area of △ADO}'}{\text{Area of △ACO}} = \dfrac{\text{AO}'^2}{\text{AO}^2} \\[1em] = \dfrac{r^2}{(3r)^2} \\[1em] = \dfrac{r^2}{9r^2} \\[1em] = \dfrac{1}{9}.$

Hence, the value of $\dfrac{\text{Area of △ADO'}}{\text{Area of △ACO}} = \dfrac{1}{9}.$