Question

In the figure, two equal circles, with centres O and O', touch each other at X.OO' produced meets the circle with centre O' at A. AC is tangent to the circle with centre O, at the point C. O'D is perpendicular to AC. Find the value of $\frac{DO'}{CO}$

Answer 1

Answer:

1/3

Step-by-step explanation:

circles have centers O and O'.

Join OO' and produce to point A on circumference

Let their radius be p

Since they are equal circles, their radii will be equal

⇒OX=XO'=O'A = p

Given that O'D is perpendicular to AC

⇒∠O'DA = 90°

We know radius ⊥ tangent at point of contact

∴ OC⊥AC ⇒ ∠OCA = 90°

In ΔO'DA and ΔOCA

∠A is common

∠OCA = ∠O'DA = 90°

∴By AA criterion, ΔOCA is similar to ΔO'DA

⇒ OC / O'D = OA / O'A

OA = p+2p = 3p

O'A = p

∴OA / O'A = 3p / p = 3/1

∴OC / O'D = 3/1

⇒ DO' / CO = 1/3