Assertion (A)
If two tangents are drawn to a circle from an external point, they subtend equal angles at the centre.
Reason (R)
A parallelogram circumscribing a circle is a rhombus.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R)is true.
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
Assertion :-
We know that if two tangents are drawn to a circle from an external point, they subtend equal angles at the centre.
Reason:-
Given, a parallelogram ABCD circumscribes a circle with centre O.
$A B=B C=C D=A D$
We know that the tangents drawn from an external point to circle are equal .
$\therefore A P=A S \ldots \ldots \ldots(i) \quad[$ tangents from $A]$
$B P=B Q \ldots \ldots \ldots .(i i) \quad[$ tangents from $B]$
$C R=C Q \ldots \ldots \ldots$ (iii) $\quad[$ tangents from $C]$
$D R=D S \ldots \ldots \ldots \ldots(i v) \quad[$ tangents from $D]$
$\therefore A B+C D=A P+B P+C R+D R$
$=A S+B Q+C Q+D S[$ from $(i),(i i),(i i i)$ and $(i v)]$
$=(A S+D S)+(B Q+C Q)$
$=A D+B C$
Thus, $(A B+C D)=(A D+B C)$
$\Rightarrow 2 A B=2 A D[\because$ opposite sides of a parallelogram are equal $]$
$\Rightarrow A B=A D$
$\therefore C D=A B=A D=B C$
Hence, ABCD is a rhombus.