NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 9: Circles - Exercise 9.4

Question 11

Step 1: We are given that:

• O is the centre of the circle.
• The radius of the circle is \(OE = 5\,\text{cm}\).
• The distance from O to point T is \(OT = 13\,\text{cm}\).
• AB is the tangent at point E, and tangents are always perpendicular to the radius at the point of contact.

Step 2: In right-angled triangle OEA:

• Hypotenuse = OT = \(13\,\text{cm}\)
• One side = OE (radius) = \(5\,\text{cm}\)
• The other side = AE (which is half of AB, since AB is symmetric).

Step 3: By Pythagoras’ theorem:

\[ OT^2 = OE^2 + AE^2 \]

Substituting the values:

\[ 13^2 = 5^2 + AE^2 \]

\[ 169 = 25 + AE^2 \]

Step 4: Simplify to find AE:

\[ AE^2 = 169 - 25 = 144 \]

\[ AE = \sqrt{144} = 12\,\text{cm} \]

Step 5: Since AB is tangent at E and passes through both sides of AE, the length of AB = AE = \(12\,\text{cm}\).

Final Answer: The length of the tangent AB is:

\[ AB = 12\,\text{cm} \]