Co-ordinate Geometry - circles

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Shaswata Roy ·2013-05-22 11:06:42

\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}

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rahul ·2013-05-23 02:46:34

Nice.. btw what if one didn't knw this relation??

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2305

Shaswata Roy ·2013-05-23 03:48:11

Then in that case:

Assume that A and B are (r cosÎ±,r sinÎ±) and (r cosÎ²,r sinÎ²). 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

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circles - nice problem

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circles - nice problem

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