Question
A point P is 13 cm away from the centre O of a circle of radius 5 cm. Find the length of the tangent drawn from P to the circle.
Solution — Step by Step
Draw a radius OQ to the point of tangency Q, and connect OP and PQ. The key theorem here: a tangent to a circle is always perpendicular to the radius at the point of tangency. So angle OQP = 90°.
We now have a right triangle OQP with the right angle at Q.
- Hypotenuse = OP = 13 cm (distance from external point to centre)
- One leg = OQ = 5 cm (radius)
- Other leg = PQ = ? (tangent length we want)
Since angle Q = 90°:
The tangent length is 12 cm.
Why This Works
The perpendicularity of radius and tangent is not just a random fact — it comes from the shortest-distance property. The radius to any point on the circle is the shortest path from the centre to that point. The tangent line just barely touches the circle, meaning the radius to the contact point must be perpendicular to it. Any tilt would create a shorter path, which would put the line inside the circle.
This perpendicularity is what gives us the right angle at Q, and once we have that, Pythagoras becomes the natural tool. We know two sides of the right triangle (OP and OQ), so finding the third is straightforward.
Notice the numbers: 5, 12, 13 is a Pythagorean triple — one of the standard ones alongside 3-4-5 and 8-15-17. NCERT and CBSE board papers love using these triples so the arithmetic stays clean.
Alternative Method — Using the Tangent-Secant Relationship
If you’ve studied power of a point, there’s an elegant one-line version.
The power of point P with respect to the circle is:
For a tangent from P, the power also equals . So:
This is the same calculation — just framed as a formula rather than drawing the triangle explicitly. For board exams, showing the triangle and using Pythagoras step by step is safer since it demonstrates understanding clearly.
Common Mistake
Many students write — adding instead of subtracting. This happens when they forget which side is the hypotenuse. The hypotenuse is always the longest side, opposite the right angle. Here, OP = 13 cm is the hypotenuse (it connects the external point to the centre, passing outside the circle). OQ and PQ are the two legs. So you subtract: .
Whenever you see a tangent problem with “distance from external point to centre” and “radius” given — immediately think 5-12-13 or 3-4-5 family. Check if the numbers fit a Pythagorean triple before grinding through the arithmetic. In this case, 5 and 13 are two sides of the 5-12-13 triple, so the answer is 12 without any real calculation needed.