Question

Two radii of a circle are inclined at ${130}^{\circ }$. Tangents are drawn at the end points of the diameters formed from the radii to form a quadrilateral. Explain the special quadrilateral formed.

Answer 1

Consider a circle having the centre at O.
Let OA and OB are the 2 radii of the given circle.
Let $\angle AOB=1300.$
Now, produce AO to X, such that AX is a diameter.
Also produce BO to Y, such that BY is a diameter.
Now, we will draw tangents at the end point A of diameter AX and end point B of diameter BY
Suppose the 2 tangents meet at point P outside the circle.
Now, AOBP becomes a quadrilateral.
We know that tangent drawn to a circle is always $\perp$ to its radius at the point of contact.
Hence, $OA\perp APandOB\perp BP.$
$\Rightarrow \angle OAP=\angle OBP={90}^{\circ }$
In quadrilateral AOBP,
$\angle APB+\angle OBP+\angle AOB+\angle AOP=3600$[Angle sum property]
$\Rightarrow \angle APB+900+1300+900=1800$
$\Rightarrow \angle APB=500$
Now, in quadrilateral AOBP
$\angle OAP+\angle OBP=900+900=1800$
$\angle APB+\angle AOB=500+1300=1800$
Now, we know that in a quadrilateral, if both the pairs of opposite angles are supplementary, then
Quadrilateral is a cyclic quadrilateral
$\Rightarrow AOBP$ is a cyclic quadrilateral