2305\mathcal{PA\cdot PB=PT}^{2}
Where PT is the length of the tangent drawn from P to the circle.Here O (0,0) is the centre of the circle.
\mathcal{PT}^{2}\mathcal{=OP}^{2}-r^{2}=m^{2}+n^{2}-r^{2}
A line is drawn through a fixed point P (m,n) to the circle x2 + y2 = r2 at A and B. Find PA x PB.
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3 Answers
2305
2305Then in that case:
Assume that A and B are (r cosα,r sinα) and (r cosβ,r sinβ). Find the equation of the line.Equate the slopes.Use distance formula and hope that the answer comes in terms of m,n and r.
It would be too long!
Anyways there's another way to do the problem.
PAxPB = PCxPD (where COD is a diameter of the circle)
=(OP+r)(OP-r)
=OP2-r2 = m2+n2-r2