O is the centre of a circle. A point P lies outside the circle such that OP = 13 cm and the radius of the circle is 5 cm. A tangent from P touches the circle at T. What is the length of the tangent segment PT (in cm)?

Introduction / Context:
This question is another classic application of the tangent from an external point to a circle. When you draw a tangent from a point outside a circle, the segment from the point of tangency to the centre is perpendicular to the tangent. This leads to a right triangle where the distance from the external point to the centre is the hypotenuse, the radius is one leg, and the tangent segment is the other leg. The Pythagoras theorem then gives a direct way to compute the length of the tangent segment. Such problems are widely used in aptitude tests to combine circle geometry with right triangle calculations.

Given Data / Assumptions:
• O is the centre of the circle.
• P is an external point such that OP = 13 cm.
• Radius of the circle, OT, is 5 cm.
• PT is a tangent from P to the circle, touching it at T.
• OT is perpendicular to PT at T, forming a right angle at T.
• We need to find the length of segment PT in centimetres.

Concept / Approach:
In triangle OPT, OT is the radius and PT is a tangent, so angle ∠OTP is 90 degrees. Therefore, triangle OPT is right angled at T. In any right triangle, the Pythagoras theorem states that the square of the hypotenuse equals the sum of the squares of the other two sides. Here, OP is the hypotenuse, and OT and PT are the legs. Hence, OP^2 = OT^2 + PT^2. Rearranging gives PT^2 = OP^2 - OT^2, and taking the square root gives PT. This is a direct and efficient way to compute the length of the tangent segment.

Step-by-Step Solution:
Step 1: Recognise that OT ⟂ PT, so triangle OPT is a right triangle with right angle at T. Step 2: Use the Pythagoras theorem: OP^2 = OT^2 + PT^2. Step 3: Substitute OP = 13 cm and OT = 5 cm into the equation: 13^2 = 5^2 + PT^2. Step 4: Compute the squares: 13^2 = 169 and 5^2 = 25, so 169 = 25 + PT^2. Step 5: Rearrange to solve for PT^2: PT^2 = 169 - 25 = 144. Step 6: Take the square root: PT = √144 = 12 cm.

Verification / Alternative check:
We can quickly verify by checking that OT^2 + PT^2 equals OP^2: 5^2 + 12^2 = 25 + 144 = 169, which matches 13^2. This is the well known 5, 12, 13 Pythagorean triple, which indicates our calculation is correct. The tangent length must be less than the distance OP and greater than the radius, and 12 cm satisfies that expectation, lying between 5 cm and 13 cm, so the result is numerically reasonable as well.

Why Other Options Are Wrong:
Option √194 cm would give OT^2 + PT^2 = 25 + 194 = 219, which does not equal 169, so it is inconsistent with the Pythagoras theorem. Option 10 cm gives 25 + 100 = 125, again not equal to 169. Option 8 cm gives 25 + 64 = 89, still not 169. Therefore, all of these violate the basic right triangle relationship between OP, OT and PT in this configuration.

Common Pitfalls:
Students sometimes incorrectly treat OP as a leg rather than the hypotenuse or forget that the radius meets the tangent at a right angle. Another common error is to subtract in the wrong order, computing 5^2 - 13^2 instead of 13^2 - 5^2, which would yield a negative number under the square root. To avoid these mistakes, always identify the hypotenuse as the longest side opposite the right angle and write the Pythagoras formula with the hypotenuse on the left side: hypotenuse^2 = leg1^2 + leg2^2.

Final Answer:
The length of the tangent PT is 12 cm.