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Two circles, each of radius 44 cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then the radius of the third circle, in cm, is

Solution

✅ Correct Option: 2

Let's break down this step-by-step to understand how three circles with a common tangent create a beautiful geometric relationship.


We have:

Two circles of radius 4 cm each that touch externally

A third circle of radius r cm that touches both circles externally

All three circles share a common tangent

The key insight: When three circles have a common tangent, their centers form a right triangle.


When a line is tangent to a circle, it's perpendicular to the radius at the point of tangency.

Since we have a common tangent touching all three circles, the line from each circle's center to the tangent point is perpendicular to this tangent line. This creates our right triangle with the right angle at the point where the tangent touches the middle circle.

Let's call the centers:

Solution figure for CAT 2019 QA question 19 (Geometry)

A: Center of first circle (radius 4 cm)

B: Center of second circle (radius 4 cm)

P: Center of third circle (radius r cm)

The distances between centers are:

AB = 8 cm (since two circles of radius 4 each touch externally: 4+4=84 + 4 = 8)

AP = 4+r4 + r (since circles touch externally: sum of their radii)

BP = 4−r4 - r (this comes from the common tangent condition)

Why is BP = 4−r4 - r and not 4+r4 + r? This happens because for the common tangent to exist, the third circle must be positioned such that it creates this specific geometric relationship. The common tangent forces this distance relationship.


In our right triangle APB (with right angle at B):

Hypotenuse = AP = 4+r4 + r

One side = AB = 8

Other side = BP = 4−r4 - r

Using Pythagoras theorem:

(4+r)2=42+(4−r)2(4 + r)^2 = 4^2 + (4 - r)^2

(4+r)2=16+(4−r)2(4 + r)^2 = 16 + (4 - r)^2

Expanding:

16+8r+r2=16+16−8r+r216 + 8r + r^2 = 16 + 16 - 8r + r^2

16+8r=32−8r16 + 8r = 32 - 8r

16r=1616r = 16

r=1r = 1 cm


The radius of the third circle is 1 cm.

This elegant result shows how the constraint of having a common tangent forces a specific geometric relationship, leading to a unique solution.

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