SOLUTION: Two circles are tangent to each other at A and the centre of the larger circle is at C. The lines AB and FC are perpendicular diameters of the larger circle. If BD = 9 cm and FE =

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Question 1150432: Two circles are tangent to each other at A and the centre of the larger circle is at C. The lines AB and FC are perpendicular diameters of the larger circle. If BD = 9 cm and FE = 5 cm, then what is the radius of the smaller circle, in centimeters?
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Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The statement of the problem is faulty; but the intent is clear. FC is a radius, not a diameter; let G be the point on the large circle that makes FG a diameter.

Let H be the other point of intersection of the small circle with diameter FG.

Let x be the length of segment CD.

Since BD = 9, the radius of the large circle is x+9. In particular, AC = 9

Since EF = 5, CE = CH = x+4..

AD and EH are chords of the small circle, intersecting at C. From the theorem about the lengths of the pieces of two intersecting chords in a circle,





The diameter of the small circle is x + (x+9) = 2x+9 = 41.

So the radius of the small circle is 20.5.


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