(i) MQBN is a rectangle, because MQ ⊥ TQ (radius ⊥ tangent) and NB ⊥ TQ (radius ⊥ tangent), so MQ ∥ NB, and MN ∥ QB. All angles are 90°.
(ii) Yes, MN ∥ PA.
Since MA ⊥ TP and NB ⊥ TQ, and TP ∥ TQ are lines from the same external point, both radii MA and NB are perpendicular to the respective tangents. By similar reasoning, MA ⊥ TP, so MN (joining centres) is parallel to the tangent segment PA.
(iii) Since TQ and TP are tangents from external point T to the larger circle:
$$TQ = TP = 80 \text{ cm}$$
$$TB = TQ - BQ = 80 - 35 = \mathbf{45 \text{ cm}}$$
(iv) In rectangle MQBN: MQ = NB + BN's projection, but using right triangle MNQ (since MN ∥ QB and MQBN is a rectangle):
$$QB = MN \text{ (opposite sides of rectangle)}$$
Wait — we use Pythagoras in △MNQ where NQ = NB + BQ is not direct. Since MQBN is a rectangle:
In right △MNQ: $MQ = 13$ cm (hypotenuse? No — MQ is a radius).
Using right △ with $MQ = 13$, $NQ = NB = 8$ (since MQBN is a rectangle, NQ = MB and MN = QB):
Actually, $MN = QB = 35$ cm from the rectangle, but we must find MN properly:
In right triangle: $MN^2 = MQ^2 - (MQ - NB)^2$...
Using right △ formed: $MN^2 + (MQ - NB)^2 = ?$
$MN^2 = BQ^2 + (MQ-NB)^2$... No.
In rectangle MQBN: $MN = BQ = 35$ cm and $QB \parallel MN$, $MQ = 13$ cm.
But since MQ = 13 is not equal to NB = 8, MQBN is a trapezium (not rectangle). So:
Draw perpendicular from N to MQ: the horizontal distance = BQ projected, vertical difference = MQ − NB = 13 − 8 = 5 cm.
$$MN^2 = BQ^2 + (MQ - NB)^2 = 35^2 + 5^2 = 1225 + 25 = 1250$$
$$MN = \sqrt{1250} = 25\sqrt{2} \approx 35.4 \text{ cm}$$
$$\boxed{MN = 25\sqrt{2} \text{ cm}}$$
Source: Circles, Chapter 10, Sections 10.2–10.3
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