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Analytical Geometry
to the circle x^2+y^2=25/2 AB is a chord of length 5cm.P is a point such that PA=4cm, PB=3cm find the distance between origin and point P.
Comments (2)
AM = MB = 5/2 cm
PA = 4 cm and PB = 3 cm => PAB is a rt. triangle
in tri.OAM, OM2 = OA2 - AM2 => OM2 = 25/2 - 25/4 = 25/4 => OA = 5/2
=> AM = OM => <OAM = 450
using triangle PAQ and PBQ and applying pytha thm we can find the value of x and PQ, where MQ = x
on solving x = 7/10 hence PQ = 13/5
from the triangle PAQ, Tan(450 + @) = PQ/AQ = (13/5) / (16/5) = 13/16 => Tan@ = 11/21 => Cos@ = 21/sqrt.310
in tr.OAP, COS@ = (OA2 + AP2 - OP2) / 2.OA.AP
now OP can easily be calculated
plz ignore the calculation mistake if done anywhere by mistake
but i think the logic is clear to u
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see d fig......
noow using cosine formula in AOP we get
cos(45+theta)=(25/2)^2+9-x^2/15rt2
-1/3=25/2+9-x^2
x=rt131/6