Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

CBSE - X - Mathematics - Circles

if PT is a tangent and PAB is a secant to a circle from an external point P,then show that PTsquare = PA*PB

Asked by 23rd February 2013, 6:04 AM
Answered by Expert

Answer:

Given : PAB is secant intersecting the circle with centre O at A and B and a tangent PT at T.
 
Draw OM perpendicular to AB and join OA, OB and OP
 
Since, OM is perpendicular to AB and a line drawn through the center is perpendicular to the chord, it bisects it also. (Can be proven easily by RHS congruency)
 
Therefore, AM = MB. 
 
Now, PA*PB = (PM-AM) * (PM+MB)
Since AM = MB
Therefore, PA*PB = (PM-AM) * (PM+AM)
PA*PB = PM2-AM2 (1)
 
Also, since triangle (PMO) is a right angled triangle, therefore, by pythagoras theorem, 
PM2 = OP2 - OM2
Also, triangle AMO is right angles triangle, therefore, AM2 = OA2 - OM2
 
Substituing it back in (1)
 
PA*PB = OP2 - OM2 - (OA2 - OM2)
PA*PB = OP2 - OA2
PA*PB = OP2 - OT2 (OA = OT as it is the radius of the circle)
PA*PB = PT2 (By using pythagoras theorm in triangle POT)
 
Hence, proved
Answered by Expert 23rd February 2013, 8:34 PM

Rate this answer

  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Report an issue
Your answer has been posted successfully!