Question

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

Answer 1

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

Construction: Join OB and OC.
Draw OA || XY

Now, XB || AO
$\Rightarrow \angle XBO+\angle AOB={180}^{\circ }$ (sum of adjacent interior angles is ${180}^{\circ }$)
Now, $\angle XBO={90}^{\circ }$ (A tangent to a circle is perpendicular to the radius through the point of contact)
$\Rightarrow {90}^{\circ }+\angle AOB={180}^{\circ }$
$\Rightarrow \angle AOB={180}^{\circ }-{90}^{\circ }={90}^{\circ }$
Similarly, $\angle AOC={90}^{\circ }$
$\angle AOB+\angle AOC={90}^{\circ }+{90}^{\circ }={180}^{\circ }$

Hence, BOC is a straight line passing through O.

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