Lengths of Tangents from External Point (Prove) MCQs Quiz | Class 10

Understanding Tangents from an External Point

In geometry, a tangent to a circle is a straight line that touches the circle at exactly one point, called the point of contact. This quiz focused on a fundamental theorem related to tangents: the lengths of tangents drawn from an external point to a circle are equal.

Key Concepts and Properties of Tangents:

• A tangent to a circle is always perpendicular to the radius drawn to the point of contact. This forms a right angle (90 degrees) at the point where the tangent meets the circle.
• From any point outside a circle, it is possible to draw exactly two tangents to that circle.
• The lengths of these two tangents drawn from the same external point to the circle are always equal. This is a crucial property for solving many problems.
• The line segment joining the external point to the center of the circle bisects the angle between the two tangents.
• The line segment joining the external point to the center of the circle also bisects the angle between the two radii at the center that connect to the points of contact.

Proof: Two tangents from external point are equal

Let’s consider a circle with center O and an external point P. Let PA and PB be two tangents drawn from P to the circle, with A and B as the points of contact on the circle.

To prove that PA = PB, we follow these steps:

1. Construction: Join OA, OB, and OP.
2. Radius-Tangent Property: We know that the radius is perpendicular to the tangent at the point of contact. Therefore:
   • Angle OAP = 90 degrees (OA is radius, PA is tangent)
   • Angle OBP = 90 degrees (OB is radius, PB is tangent)
3. Consider Triangles OAP and OBP:
   • OA = OB (Both are radii of the same circle)
   • OP = OP (Common side to both triangles)
   • Angle OAP = Angle OBP = 90 degrees (From the radius-tangent property)
4. Congruence: By the RHS (Right angle – Hypotenuse – Side) congruence criterion, Triangle OAP is congruent to Triangle OBP. (R is for the right angle at A and B, H is for the common hypotenuse OP, and S is for the equal radii OA and OB).
5. Conclusion: Since the two triangles are congruent, their corresponding parts are equal (CPCTC – Corresponding Parts of Congruent Triangles are Congruent). Therefore, PA = PB.

This proof is a fundamental concept in Class 10 Mathematics and often appears in examinations.

Quick Revision Points:

• A line segment that touches a circle at only one point is called a tangent.
• The radius drawn to the point of contact is always perpendicular to the tangent.
• The lengths of two tangents drawn from an external point to a circle are always equal.
• The RHS congruence criterion (Right Angle, Hypotenuse, Side) is key to proving this theorem.

Extra Practice Questions:

1. A point Q is 25 cm from the center of a circle. The radius of the circle is 7 cm. Find the length of the tangent from Q to the circle.
2. If tangents PA and PB from an external point P to a circle with center O are inclined to each other at an angle of 70 degrees, find the measure of angle AOB.
3. A quadrilateral PQRS is drawn to circumscribe a circle. Prove that PQ + RS = PS + QR.
4. In the figure, if PA and PB are tangents from an external point P to a circle with center O and angle OPA = 30 degrees, find the length of the radius if OP = 10 cm.
5. Two tangents PA and PB are drawn to a circle with center O from an external point P. If PA = 8 cm and angle AOP = 60 degrees, find the length of OA.

Practicing these concepts will strengthen your understanding of tangents and their properties in circle geometry.