Prove that the line segment joining the point of contact of two parallel tangents of a circle passes through its center.

Given: Two parallel tangents of a circle and a line segment joining joining the point of contact of the given tangents.

To do: To prove that the this line segment passes through the center.

Solution:

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

Let us Join OB and OC.

Now, $XB\parallel AO$

We know the sum of adjacent interior angles is $180^{o}$.

$\therefore \ \angle XBO+\angle AOB=180^{o}$

$\angle XBO=90^{o} \ (A\ tangent\ to\ a\ circle\ is\ perpendicular\ to\ the\ radius\ through\ the\ point\ of\ contact)$

$\Rightarrow 90^{o} +\angle AOB=180^{o}$

$\Rightarrow \angle AOB=90^{o}$

Similarly, $\angle AOC=90^{o} $

$\angle BOC=\angle AOB+\angle AOC=90^{o}+90^{o}=180^{o}$

Thus we find that $\angle BOC$ is a straight line passing through the center O.

Or we can say that BC is a straight line passing through the center O.

Hence, it is proved that a line segment joining the points of contact B and C of the tangents of the circle passes through the circle. 

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Updated on 10-Oct-2022 10:20:11

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