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Constructions

Question
CBSEENMA10008471
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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.

Solution

Given : PA and PB arc two tangents drawn from an external point P to a circle with centre O.
To prove : ∠AOB + ∠APB = 180°
Const : Join OA and OB.
Proof : ∵ The tangent at any point of circle is perpendicular to the radius through the point of contact.
∴ ∠OAP = 90°    .....(i)
and    ∠OBP = 90°    .....(ii)
Adding (i) and (ii), we get
∠OAP + ∠OBP = 180°
Now in quadrilateral AOBP,
∠OAP + ∠OBP + ∠APB + ∠AOB = 360°
⇒    180° + ∠APB + ∠AOB = 360°
∴ ∠APB + ∠AOB = 180°.

Some More Questions From Constructions Chapter

A tangent to a circle intersects it in ___________ point(s).

A line intersecting a circle in two points is called a ____________.

A circle many have ______ parallel tangents.

The common point of a tangent and the circle is called ____________.

In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠ POQ = 110°, then ∠ PTQ is equal to
(A) 60°    (B) 70°
(C) 80°    (D) 90°



Fig. 10.11

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠ POA is equal to

(A) 50°    (B) 60°
(C) 70°    (D) 80°.

Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

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