Geometry-11 (CAT-2002)

There is a common chord of two circles with radius 15 and 20.The distance between the two centres is 25.The length of the chord is:
(1)48   (2)24   (3)36   (4)28

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Solution:

“For intersecting circles, the line joining the centres is perpendicular to and bisects the common chord”

P₀P₂ = a;  P₁P₂ = b;

P₁P₂ = h => length of the chord = 2h

Radius of bigger circle = r_o = 20

Radius of smaller circle = r₁ = 15

Applying Pythagoras’ to the right triangle P₀P₂P₄,

a²+h² = r₀² = 20² = 400   ---(1)

Applying Pythagoras’ to the right triangle P₁P₂P₄,

b²+h² = r₁² = 15² = 225   ---(2)

(1)-(2) gives,

a²-b² = 175        ---(3)

a+b = distance between the centres = 25   ----(4)

(3)/(4) gives:

a-b = 7                ---(5)

Solving (4) and (5), a = 16; b = 9;

Substituting ‘a’ value in (1),

16²+h² = 400 => h² = 400-256 = 144 => h =12

Hence length of chord = 2h = 24

Answer (2)