Question 9 of 11intermediate🔍 AnalyzeLong Answer4 marks

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.

Correct Answer

Let us consider a circle with centre O. Let P be an external point from which two tangents PQ and PR are drawn, touching the circle at points Q and R respectively. We need to prove that the angle between the tangents, ∠QPR\angle QPR, and the angle subtended by the line-segment joining the points of contact at the centre, ∠QOR\angle QOR, are supplementary.

We join the centre O to the external point P and the points of contact Q and R. According to the property of tangents, the angle between the radius and the tangent at the point of contact is a right angle. Therefore, ∠OQP=90∘\angle OQP = 90^\circ and ∠ORP=90∘\angle ORP = 90^\circ.

In quadrilateral OQPR, the sum of all interior angles is 360∘360^\circ. We can write this as:

∠OQP+∠QPR+∠PRO+∠QOR=360∘\angle OQP + \angle QPR + \angle PRO + \angle QOR = 360^\circ

Substituting the values of the right angles:

90∘+∠QPR+90∘+∠QOR=360∘90^\circ + \angle QPR + 90^\circ + \angle QOR = 360^\circ

Simplifying the equation, we get:

∠QPR+∠QOR+180∘=360∘\angle QPR + \angle QOR + 180^\circ = 360^\circ ∠QPR+∠QOR=180∘\angle QPR + \angle QOR = 180^\circ

Hence, it is proved that the angle between the two tangents is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Exercise: EXERCISE 10.2 | Q: 10 | (Page 9)
For More Understanding

Explanation

The question asks for a geometric proof based on the properties of tangents to a circle. The provided context, specifically the proof of Theorem 10.2 and Example 2, establishes that the radius is perpendicular to the tangent at the point of contact.

Using this fact, the solution constructs a quadrilateral formed by the centre, the external point, and the points of contact. By applying the angle sum property of a quadrilateral (360∘360^\circ), the relationship between the angles is derived.

This approach is standard for Class 10 Geometry and aligns perfectly with the textbook context provided.

Solution Steps

  1. Step 1: Draw a circle with centre O and let P be an external point. Draw tangents PQ and PR touching the circle at Q and R.

  2. Step 2: Join O to P, Q, and R. Note that ∠OQP=90∘\angle OQP = 90^\circ and ∠ORP=90∘\angle ORP = 90^\circ (radius is perpendicular to tangent).

  3. Step 3: Consider quadrilateral OQPR. The sum of angles in a quadrilateral is 360∘360^\circ.

  4. Step 4: Write the angle sum equation:

    ∠OQP+∠QPR+∠PRO+∠QOR=360∘\angle OQP + \angle QPR + \angle PRO + \angle QOR = 360^\circ

  5. Step 5: Substitute 90∘90^\circ for ∠OQP\angle OQP and ∠ORP\angle ORP to get:

    180∘+∠QPR+∠QOR=360∘180^\circ + \angle QPR + \angle QOR = 360^\circ

  6. Step 6: Conclude that:

    ∠QPR+∠QOR=180∘\angle QPR + \angle QOR = 180^\circ

    ∴ The angles are supplementary.

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