In the given figure, \(TP\) and \(TQ\) are tangents to a circle with centre \(M\), touching another circle with centre \(N\) at \(A\) and \(B\) respectively. It is given that \(MQ = 13 \text{ cm}\), \(NB = 8 \text{ cm}\), \(BQ = 35 \text{ cm}\) and \(TP = 80 \text{ cm}\).
(i) Name the quadrilateral MQBN. (1)
(ii) Is MN parallel to PA? Justify your answer. (1)
(iii) Find length TB. (1)
(iv) Find length MN. (2)
Show Hint
When finding the distance between centers in a configuration with parallel radii, always construct a right triangle by drawing a perpendicular from the smaller radius to the larger one. This allows the use of the Pythagoras theorem.
Step 1: Using Properties of Tangents and Circles:
• Tangents from an external point to a circle are equal in length.
• Radius is perpendicular to the tangent at the point of contact.
(i) Nature of Quadrilateral MQBN:
MQ ⟂ TQ (radius ⟂ tangent)
NB ⟂ TQ (radius ⟂ tangent)
Since both MQ and NB are perpendicular to the same line TQ,
MQ ∥ NB.
A quadrilateral with one pair of opposite sides parallel is a Trapezium.
(ii) Checking Parallelism of MN and PA:
MN joins the centres of the circles.
PA lies along the tangents.
There is no geometric property establishing MN ∥ PA.
Therefore, MN is not parallel to PA.
(iii) Finding TB:
From tangent property:
TP = TQ = 80 cm
Now,
TQ = TB + BQ
80 = TB + 35
TB = 80 − 35
TB = 45 cm
(iv) Finding Distance Between Centres MN:
We know MQ ∥ NB.
Draw perpendicular from N to MQ meeting it at X.
Given:
MQ = 13 cm
NB = 8 cm
BQ = 35 cm
Since QX = NB = 8 cm,
MX = MQ − QX
MX = 13 − 8
MX = 5 cm
Also,
NX = BQ = 35 cm
Now in right triangle MXN:
MN² = MX² + NX²
= 5² + 35²
= 25 + 1225
= 1250
MN = √1250
= 25√2
≈ 35.36 cm
Final Answers:
(i) MQBN is a Trapezium.
(ii) MN is not parallel to PA.
(iii) TB = 45 cm.
(iv) MN = 25√2 cm (≈ 35.36 cm).
