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Diameter of a Circle

  • #1
Two chords of a circle, AB and CD, intersect at X, such that AX=BX= 6 cm. CX=4 cm and DX=9 cm. Calculate the radius of the circle, given that it has an integer value. I constructed a secant EG of a diameter FG of the circle and drew a tangent from C to it at E, then we have from the intersecting secant theroem: EFxEG=EC^2.Many thanks guys.
 

Answers and Replies

  • #2
AKG
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I don't understand your work.
 
  • #3
All I did was to draw a tangent at point C, to the circle, then drew a diameter FG of the circle such that, when you produce the diameter GF it meets the tangent at E.Sorry for the long delay in reply. Many thanks.
 
  • #4
AKG
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Is FG just any diamater that isn't parallel to the tangent at C? And I don't see why you did what you did.
 
  • #5
The reason why I have done what I did, was that this question is posed in the section of a book called "intersecting chord and secant theorems".Even though I would need to know the length EF and the tangent |EC|. Since I don't the solution to the problem eludes me.Maybe someone could start me off!
 
  • #6
Gokul43201
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The radius is quite guessable from observing that the lengths of the 4 segments satisfy Archimedis' lemma for perpendicular chords.
 
  • #7
AKG
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The chords aren't necessarily perpendicular (are they?).
 
  • #8
No, the chords aren't necessarily perpendicular. The answer is 10cm. The question is how did they get it!
 
  • #9
StatusX
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I don't think there's enough information here. Take any circle of sufficiently large size (this will be clear in a minute). Draw a chord of length 12, and call the endpoints A and B. Then draw a line from the center of this chord X, to a point C on on the arc between A and B, such that the length of CX is 4. Note this is possible since the maximum length such a segment could have is clearly 6, and the minimum can be made arbitrarily small by increasing the radius of the circle. Now extend this line to intersect the circle again at a point D, and since AX*BX=CX*DX, we must have DX=9. Am I misssing something, or is this exactly the setup you've described? Maybe you're asked to find the smallest such circle?
 
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  • #10
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StatusX said:
I don't think there's enough information here. Take any circle of sufficiently large size (this will be clear in a minute). Draw a chord of length 12, and call the endpoints A and B. Then draw a line from the center of this chord X, to a point C on on the arc between A and B, such that the length of CX is 4. Note this is possible since the maximum length such a segment could have is clearly 6, and the minimum can be made arbitrarily small by increasing the radius of the circle. Now extend this line to intersect the circle again at a point D, and since AX*BX=CX*DX, we must have DX=9. Am I misssing something, or is this exactly the setup you've described? Maybe you're asked to find the smallest such circle?
But then the smallest such circle would be one in which the longest chord is diameter, andthe radius would be equal to 6.5.

I agree, there is not enough information.
 

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