Here, length of tangent (AB) = 4 cm
OA = 5 cm
Find : radius (r) = ?
∴ OB ⊥ AB(line through centre to the point of contact is perpendicular)
Now, in right ΔOAB,
OA² = OB² + A B² (Using Pythagoras theorem]
⇒    (5)² = r² + 4²
⇒    25 = r² + 16
⇒    r² = 25 – 16
⇒ r² = 9
⇒    r = 3 cm
Hence, radius of the circle is 3 cm.

Distance between the two parallel tangent to a circle
= diameter
= 2 x r
= 2 x 4 = 8 cm.

∠OQT = 90° (line drawn through the centre to the point of contact is perpendicular to it)
∠OPT = 90° (line drawn through the centre to point of contact is perpendicular to it)
∠QTP = 100    (given)
Now, in quadrilateral OPTQ,
∠OPT + ∠OQT + ∠QTP + ∠POQ = 360°
⇒ 90° + 90° + 100° + ∠POQ = 360°
⇒ ∠POQ = 360° – 280°
⇒    ∠POQ = 80°.

Since, the tangent at any point of a crcle is perpenedicular the radius throgh the point of contact.
∴ ∠OAT = 90°
and ∠OBT = 90°
If two tangents are drawn to a circle from an external point, then they are equally inclined to the segment joining the centre to the point
∴ ∠ATO = ∠BTO
But ∠ATO = 40°
Now, ∠BTO = 40°
In quadrilateral, ∠OBT, we have
∠OAT + ∠OBT + ∠ATB + ∠AOB= 360°
⇒ 90 + 90 + 80 + ∠AOB = 360°
⇒ 260 + ∠AOB = 360°
⇒ ∠AOB = 360° – 260° = 100°

Given : A circle with centre O, an external point P and two tangents PA and PB to the circle, where A and B are points of contact (Fig. 10.36A).
To Prove : ∠APB 2∠OAB
Proof : Let ∠APB = 9 ∴ PA = PB
[∴ Tangents from an ext. pt. are equal]

   APB is ab isosceles triangle




Hence, ∠APB = 2 ∠OAB. Proved.