Q. 144.4( 21 Votes )

In two concentric

Answer :

We are given two concentric circles C1 and C2 with centre O and a chord AB of the larger circle C1 which touches the smaller circle C2 at the point P (see Fig. 10.8).

We need to prove that AP = BP.

Let us join OP. Then, AB is a tangent to C2 at P and OP is its radius. Therefore,

OPAB (Since Tangent at any point of circle is perpendicular to the radius through point of contact)

Now AB is a chord of the circle C1 and OPAB. Therefore, OP is the bisector of the chord AB, as the perpendicular from the centre bisects the chord,

Hence AP = BP

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