Prove that the line segment joining the points of contact of two parallel tangents of a circle

>

Prove that the line segment joining the point of contact of two parallel tangents to a circle is a diameter of the circle.

Solution

Given : CD and EF are two parallel tangents at the points A and B of a circle with center O.

To prove : AOB is a diameter of the circle

Construction : Join OA and OB

Draw OG | | CD

Proof : OG | | CD and AO cuts them .

⇒ ${90}^{\circ }$ + GOA = ${180}^{\circ }$ [ OA is perpendicular to CD ]

⇒ GOA = ${90}^{\circ }$)

Similarly, GOB = ${90}^{\circ }$;

Therefore, GOA + GOB = (${90}^{\circ }$ + ${90}^{\circ }$) = ${180}^{\circ }$)

=> AOB is a straight line

Hence, AOB is a diameter of the circle with center O.

>

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

Solution

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.

NCERT Previous Years Papers