Question

prove that the line segment joining the points of contact of 2 parallel tangents of a circle passes through the centre

Answer 1

Let the center of the circle be O.  Let there be two tangents T1 and T2 touching the circle at P and Q. Let T1 and T2 be parallel.

Join PO and OQ.  We know PO ⊥ T1   and  OQ ⊥ T2 (radius & tangent).

Hence,  OQ ⊥ T1  (as T1 || T2).

Hence,   PO || OQ .   Both have common point O.  Hence  POQ is a single straight line.

Hence proved.

Answer 2

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Let, XBY and PCQ be two parallel tangents to a circle with centre O.

Construction :

Join OB and OC

Draw OA || XY

Proof :

Now,

XB || AO

⟹ ∠ XBO + ∠ AOB = 180° ─━⧼ sum of adjacent interior angles is 180°

Now,

∠ XBO = 90° ─━⧼ a tangent to a circle is perpendicular to the radius through a point of contact

⟹ 90° + ∠ AOB = 180°

∴ ∠ AOB = 90°

-lly, ∠ AOC = 90°

Hence,

∠ AOB + ∠ AOC = 90° + 90° = 180°

Hence, BOC is a straight line passing through O.

Thus, line segment joining the points of contact of two parallel tangents of a circle passes through its centre.

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