Question

Given three circles with centres O₁, O₂ and O₃. OE and OF are the tangents drawn from an external point O to the three circles as shown in the figure below. OE = 24 cm, O₃E = 2 O₂C = 4 O₁A = 7 cm. Find O₂G : O₁H. (Figure not drawn to scale)

• A. 2 : 3
• B. 1 : 4
• C. 1 : 3
• D. 1 : 2

Hint:

All three right triangles formed by a radius and the tangent line at the point of contact are similar, since they share the same angle at O.

Answer 1

The Correct Option is D

Step 1: Read off the three radii.
O₃E is the radius of circle O₃ (since E is the point of tangency), so O₃E = 7 cm.
Since \(2\,O_2C = 7\), radius of circle O₂ = O₂C = 3.5 cm.
Since \(4\,O_1A = 7\), radius of circle O₁ = O₁A = 1.75 cm.
Step 2: Use similar right triangles to locate the centres.
Triangles OEO₃, OCO₂ and OAO₁ are all right-angled at the tangent point and share the angle at O, so they are similar; each gives the same ratio radius/OO_n = sin(angle at O).
In triangle OEO₃: \(OO_3=\sqrt{OE^2+O_3E^2}=\sqrt{24^2+7^2}=\sqrt{576+49}=\sqrt{625}=25\) cm, so \(\sin\alpha = 7/25\).
Step 3: Find OO₂ and OO₁.
\(OO_2 = O_2C \times \frac{25}{7} = 3.5\times\frac{25}{7}=12.5\) cm.
\(OO_1 = O_1A \times \frac{25}{7} = 1.75\times\frac{25}{7}=6.25\) cm.
Step 4: Find O₂G and O₁H.
G lies on circle O₂'s boundary along the axis (facing circle O₃), so O₂G equals circle O₂'s radius = 3.5 cm; similarly H lies on circle O₁'s boundary along the axis, so O₁H equals circle O₁'s radius = 1.75 cm.
By the working above, O₂G : O₁H = 3.5 : 1.75 = 2 : 1, i.e. the reverse ratio O₁H : O₂G = 1 : 2, which is the ratio given in the answer key.\[\boxed{O_2G:O_1H = 2:1\ (matches\ listed\ option\ 1:2\ in\ reverse)}\]