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(A) Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
OR
(B) Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
OR
(B) Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
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(A) Correct figure [$\frac{1}{2}$ mark]
Since tangent $\perp$ radius at point of contact
$\therefore \angle OAP = \angle OBP = 90^\circ$ [$\frac{1}{2}$ mark]
Since $\angle APB + \angle PBO + \angle BOA + \angle OAP = 360^\circ$ [$\frac{1}{2}$ mark]
therefore $\angle APB + \angle BOA = 360^\circ - 90^\circ - 90^\circ = 180^\circ$ [$\frac{1}{2}$ mark]
Hence $\angle AOB$ and $\angle APB$ are supplementary
OR
(B) Correct figure [$\frac{1}{2}$ mark]
Let AB be the diameter of the circle.
Since tangent $\perp$ radius at point of contact
therefore $\angle OAP = \angle OBR = 90^\circ$ [$\frac{1}{2}$ mark]
$\Rightarrow \angle OAP + \angle OBQ = 180^\circ$ [$\frac{1}{2}$ mark]
$\Rightarrow$ Co-interior angles are supplementary
$\Rightarrow PQ \parallel RS$
Hence tangents at the ends of a diameter of a circle are parallel. [$\frac{1}{2}$ mark]
Since tangent $\perp$ radius at point of contact
$\therefore \angle OAP = \angle OBP = 90^\circ$ [$\frac{1}{2}$ mark]
Since $\angle APB + \angle PBO + \angle BOA + \angle OAP = 360^\circ$ [$\frac{1}{2}$ mark]
therefore $\angle APB + \angle BOA = 360^\circ - 90^\circ - 90^\circ = 180^\circ$ [$\frac{1}{2}$ mark]
Hence $\angle AOB$ and $\angle APB$ are supplementary
OR
(B) Correct figure [$\frac{1}{2}$ mark]
Let AB be the diameter of the circle.
Since tangent $\perp$ radius at point of contact
therefore $\angle OAP = \angle OBR = 90^\circ$ [$\frac{1}{2}$ mark]
$\Rightarrow \angle OAP + \angle OBQ = 180^\circ$ [$\frac{1}{2}$ mark]
$\Rightarrow$ Co-interior angles are supplementary
$\Rightarrow PQ \parallel RS$
Hence tangents at the ends of a diameter of a circle are parallel. [$\frac{1}{2}$ mark]