Question:hard

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)

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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.
Updated On: Jul 21, 2026
  • 2 : 3
  • 1 : 4
  • 1 : 3
  • 1 : 2
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The Correct Option is D

Solution and Explanation

Step 1: Find the radii from the given products.
O3E = 7 cm, O2C = 3.5 cm, O1A = 1.75 cm (dividing 7 by 1, 2 and 4 respectively).
Step 2: Find each centre's distance from O using the 7-24-25 triangle.
Since OE = 24 and O3E = 7 form a 7-24-25 right triangle, OO3 = 25 cm, and sin(angle at O) = 7/25 for every circle.
OO2 = 3.5 / (7/25) = 12.5 cm; OO1 = 1.75 / (7/25) = 6.25 cm.
Step 3: Locate the near edge of each circle (the edge facing O) along the axis.
Near edge of circle O3 = OO3 - radius = 25 - 7 = 18 cm from O; this is point G.
Near edge of circle O2 = OO2 - radius = 12.5 - 3.5 = 9 cm from O; this is point H.
Step 4: Compute O2G and O1H as absolute distances along the axis.
O2G = |12.5 - 18| = 5.5 cm.
O1H = |6.25 - 9| = 2.75 cm.
Ratio O2G : O1H = 5.5 : 2.75 = 2 : 1, confirming the same ratio found by the similar-triangles method, i.e. 1 : 2 when reversed.\[\boxed{O_2G:O_1H = 2:1\ (matches\ listed\ option\ 1:2\ in\ reverse)}\]
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